剑指offer做题记录

1. 剑指 Offer 03. 数组中重复的数字力扣

class Solution {
public:int findRepeatNumber(vector<int>& nums) {/* 法一 排序 *//* 法二 哈希表 */map<int, int> mymap;int size = nums.size();mymap[nums[0]] = 0; for(int i= 1; i < size; ++i) {if (mymap.find(nums[i]) == mymap.end()) {mymap[nums[i]] = i;} else {return nums[i];}}return 0;/* 法3  方法3: 鸽巢原理，因为出现的元素值 < nums.size(); 所以我们可以将见到的元素 放到索引的位置，如果交换时，发现索引处已存在该元素，则重复 O(N) 空间O(1) */for(int i=0;i<nums.size();i++){while(nums[i]!=i){if(nums[nums[i]] == nums[i]) return nums[i];int tmp = nums[i];nums[i] = nums[tmp];nums[tmp] = tmp;}}return -1;}
};

2. 剑指 Offer 04. 二维数组中的查找力扣

class Solution {
public:bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {int rows = matrix.size();if (rows == 0) {return false;}int cols = matrix[0].size();if (cols == 0) {return false;}int row = 0;int col = cols - 1;while(row < rows && col >= 0) {if (matrix[row][col] == target) {return true;}else if (matrix[row][col] < target) {row++;}//这里必须加else if       否则上面row++后会越界else if (matrix[row][col] > target) {col--;}}return false;}
};

3. 剑指 Offer 05. 替换空格力扣

class Solution {
public:string replaceSpace(string s) {int len = s.size();string res;if (len == 0 || len > 10000) {return res;}int cur = 0;while(cur < len) {string tmp = "";while(cur < len && s[cur] != ' ') {tmp += s[cur];cur++;}res += tmp;if (s[cur] == ' ') {res += "%20";}cur++;}return res;}
};

4. 剑指 Offer 06. 从尾到头打印链表力扣

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:void dfs(ListNode* head, vector<int>& res) {if (!head) {return;}dfs(head -> next, res);if (head) {res.push_back(head -> val);}}vector<int> reversePrint(ListNode* head) {//法一：栈//法二：递归vector<int> res;dfs(head, res);return res;}
};

5. 剑指 Offer 09. 用两个栈实现队列力扣

两个栈，一个栈负责进入，一个栈负责出 当有元素进入，第一个栈后面加入 当有元素出队列时候，三种情况 1.两个栈都是空，则证明是空队列 2.第一个栈有元素，第二个栈没有元素，把第一栈的元素全部出栈放入第二个栈，第二个栈出栈 3.第二个栈还有元素，第二个栈出栈
class CQueue {
public:stack<int> sta1;stack<int> sta2;CQueue() {}void appendTail(int value) {sta1.push(value);}int deleteHead() {if (sta1.empty() && sta2.empty()) {return -1;}if (!sta2.empty()) {int top = sta2.top();sta2.pop();return top;} else {while(!sta1.empty()) {sta2.push(sta1.top());sta1.pop();}int top = sta2.top();sta2.pop();return top;}return -1;}
};/*** Your CQueue object will be instantiated and called as such:* CQueue* obj = new CQueue();* obj->appendTail(value);* int param_2 = obj->deleteHead();*/

6. 剑指 Offer 13. 机器人的运动范围力扣

class Solution {
public:/* 注意下这题dfs上下左右后，用不用把visit[i][j]再赋false */int CountSum(int m, int n) {int sum = 0;while(m > 0) {sum += m % 10;m /= 10;}while(n > 0) {sum += n % 10;n /= 10;}return sum;}int dfs(int i, int j, int m, int n, vector<vector<bool>>& visit, int k) {if (i < 0 || j < 0 || i >= m || j >= n || visit[i][j] == true || CountSum(i, j) > k ) {return 0;}visit[i][j] = true;return 1 + dfs(i + 1, j, m, n, visit, k) + dfs(i - 1, j, m, n, visit, k) + dfs(i, j + 1, m, n, visit, k) + dfs(i, j - 1, m, n, visit, k);}int movingCount(int m, int n, int k) {vector<vector<bool>> visit(m, vector<bool>(n, false));return dfs(0, 0, m, n, visit, k);}
};

7. 剑指 Offer 15. 二进制中1的个数力扣

class Solution {
public:int hammingWeight(uint32_t n) {//n&(-n)得到n的最后一位的1//n&(n-1) 去除 n 的位级表示中最低的那一位 1//位运算理论https://www.cnblogs.com/nobita/p/14320239.htmlint count = 0;while(n) {n = n & (n - 1);count++;}return count;}
};

8. 剑指 Offer 16. 数值的整数次方力扣

class Solution {
public://如果n是奇数，x的n/2次方乘以x的n/2次方乘以x//若n是偶数，x的n/2次方乘以x的n/2次方//若n是负数怎么办/* -2.00000 -2147483648会出错，原因是-2147483648超出int范围，可以直接在原题的函数参数double myPow(double x, int n)里的int n 改成long n   因为用例里哪怕是int类型支持转成long类型*/double myPow(double x, long n/* 题原本是int n */) {if (n == 0) {return 1.0;}if (n == 1) {return x;}if (x == 1.00) {return x;}if (n < 0) {x = 1/x;n = -n;}double res = 1.0;double tmp = myPow(x, abs(n) >> 1);if (abs(n) % 2 == 1) {res = tmp * tmp * x;} else {res = tmp * tmp;}if (n > 0) {return res;} else {return 1.0 / res;}return res;}
};

9. 剑指 Offer 21. 调整数组顺序使奇数位于偶数前面力扣

class Solution {
public:vector<int> exchange(vector<int>& nums) {int len = nums.size();if (len == 0) {return nums;}int left = 0;int right = len - 1;while(left <= right) {while(left < len && nums[left] % 2 == 1) {left++;}while(right >= 0 && nums[right] % 2 == 0) {right--;}if (left <= right) {swap(nums[left], nums[right]);left++;right--;}}return nums;}
};

10. 剑指 Offer 26. 树的子结构力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:bool compareTwoTrees(TreeNode* A, TreeNode* B) {if (B == nullptr) {return true;}if (A == nullptr) {  /*if (A == nullptr && B != nullptr)*/return false;}if (A -> val != B -> val) {return false;}return compareTwoTrees(A -> left, B -> left) && compareTwoTrees(A -> right, B -> right);}bool isSubStructure(TreeNode* A, TreeNode* B) {if (B == nullptr) {return false;}if (A == nullptr) {return false;}return isSubStructure(A -> left, B) || isSubStructure(A -> right, B) || compareTwoTrees(A, B);}
};

11. 反转链表力扣

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {/* if (head == nullptr || head -> next == nullptr) {return head;}ListNode* pre = nullptr;ListNode* cur = head;ListNode* tmp_next = nullptr;while(cur) {tmp_next = cur -> next;cur -> next = pre;pre = cur;cur = tmp_next;}return pre; */if (head == nullptr || head -> next == nullptr) {return head;}ListNode* tmp = reverseList(head -> next);head -> next -> next = head;head -> next = nullptr;return tmp;}
};

12 剑指 Offer 27. 二叉树的镜像力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:void dfs(TreeNode* &root) {if (!root || (root -> left == nullptr && root -> right == nullptr)) {return;}TreeNode* tmp = root -> left;root -> left = root -> right;root -> right = tmp;dfs(root -> left);dfs(root -> right);}TreeNode* mirrorTree(TreeNode* root) {dfs(root);return root;}
};

13 剑指 Offer 28. 对称的二叉树力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:bool dfs(TreeNode* node1, TreeNode* node2) {if (node1 == nullptr && node2 == nullptr) {return true;}if (node1 == nullptr || node2 == nullptr) {return false;}if (node1 -> val != node2 -> val) {return false;}return dfs(node1 -> left, node2 -> right) && dfs(node1 -> right, node2 -> left);}bool isSymmetric(TreeNode* root) {if (!root) {return true;}return dfs(root -> left, root -> right);}
};

14 剑指 Offer 30. 包含min函数的栈力扣

class MinStack {
public:/** initialize your data structure here. */stack<int> res;stack<int> store;MinStack() {}void push(int x) {res.push(x);if (store.empty() || store.top() > x) {store.push(x);} else {store.push(store.top());}}void pop() {res.pop();store.pop();}int top() {return res.top();}int min() {return store.top();}
};/*** Your MinStack object will be instantiated and called as such:* MinStack* obj = new MinStack();* obj->push(x);* obj->pop();* int param_3 = obj->top();* int param_4 = obj->min();*/

15 剑指 Offer 31. 栈的压入、弹出序列力扣

class Solution {
public:bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {int size1 = pushed.size();int size2 = popped.size();stack<int> sta;if (size1 != size2) {return false;}if (size1 == 0 && size2 == 0) {return true;}int i = 0;int j = 0;for (; i < size1; ++i) {sta.push(pushed[i]);while(sta.size() != 0 && sta.top() == popped[j]) {sta.pop();j++;}}return sta.empty();}
};

剑指 Offer 34. 二叉树中和为某一值的路径力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:void dfs(TreeNode* root, int sum, int target, vector<int> vec, vector<vector<int>>& res) {if (!root) {return;}sum += root -> val;vec.push_back(root -> val);if (target == sum && root -> left == nullptr && root -> right == nullptr) {res.push_back(vec);}dfs(root -> left, sum, target, vec, res);dfs(root -> right, sum, target, vec, res);vec.pop_back();sum -= root -> val;}vector<vector<int>> pathSum(TreeNode* root, int target) {int sum = 0;vector<vector<int>> res;vector<int> vec;dfs(root, sum, target, vec, res);return res;}
};

17 剑指 Offer 35. 复杂链表的复制力扣

/*
// Definition for a Node.
class Node {
public:int val;Node* next;Node* random;Node(int _val) {val = _val;next = NULL;random = NULL;}
};
*/
class Solution {
public:Node* copyRandomList(Node* head) {if(head == nullptr) return nullptr;Node* cur = head;unordered_map<Node*, Node*> map;// 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射while(cur != nullptr) {map[cur] = new Node(cur->val);cur = cur->next;}cur = head;// 4. 构建新链表的 next 和 random 指向while(cur != nullptr) {map[cur]->next = map[cur->next];map[cur]->random = map[cur->random];cur = cur->next;}// 5. 返回新链表的头节点return map[head];}
};

18 剑指 Offer 38. 字符串的排列力扣

（1）set去重

class Solution {
public:void dfs(string s, int i, int len, set<string>& mySet, string tmpStr, vector<bool>& visit) {if (i == len) {mySet.insert(tmpStr);return;}for(int j = 0; j < len; ++j) {if (visit[j] == false) {tmpStr += s[j];visit[j] = true;dfs(s, i + 1, len, mySet, tmpStr, visit);visit[j] = false;tmpStr.pop_back();}}}vector<string> permutation(string s) {set<string> mySet;string tmpStr;int len = s.size();vector<bool> visit(len, false);vector<string> res;dfs(s, 0, len, mySet, tmpStr, visit);set<string>::iterator iter = mySet.begin();for(; iter != mySet.end(); ++iter) {res.push_back(*iter);}return res;}
};

19 剑指 Offer 40. 最小的k个数力扣

class Solution {
public:vector<int> getLeastNumbers(vector<int>& arr, int k) {priority_queue<int, vector<int>, less<int>> buffer;vector<int> res;int len = arr.size();if (len == 0 || k == 0) {return res;}for(int i = 0; i < len; ++i) {if (i < k) {buffer.push(arr[i]);} else {if (arr[i] < buffer.top()) {buffer.pop();buffer.push(arr[i]);}}}while(buffer.size()!= 0) {res.push_back(buffer.top());buffer.pop();}return res;}
};

20 剑指 Offer 45. 把数组排成最小的数力扣

class Solution {
public:string minNumber(vector<int>& nums) {vector<string> strs;string res;for(auto num:nums)strs.push_back(to_string(num));sort(strs.begin(),strs.end(),compare);for(auto str:strs)res+=str;return res;}
private:static bool compare(const string &a,const string &b){return a+b<b+a;}
};

21 剑指 Offer 47. 礼物的最大价值力扣

class Solution {
public:int maxValue(vector<vector<int>>& grid) {int rows = grid.size();if (rows == 0) {return 0;}int cols = grid[0].size();vector<vector<int>> vec(rows, vector<int>(cols, 0));vec[0][0] = grid[0][0];for(int i = 1; i < rows; ++i) {vec[i][0] = vec[i - 1][0] + grid[i][0];}for(int i = 1; i < cols; ++i) {vec[0][i] = vec[0][i - 1] + grid[0][i];}for(int i = 1; i < rows; ++i) {for(int j = 1; j < cols; ++j) {vec[i][j] = grid[i][j] + max(vec[i - 1][j], vec[i][j - 1]);}}return vec[rows - 1][cols - 1];    }
};

22 剑指 Offer 48. 最长不含重复字符的子字符串力扣

class Solution {
public:int lengthOfLongestSubstring(string s) {/* 滑动窗口int len = s.size();if (len == 0 || len == 1) {return len;}int left = 0, right = left + 1;unordered_set<char> mySet;int resLen = 1;int tmplen = 1;mySet.insert(s[left]);while(right < len) {while(left < right && mySet.find(s[right]) != mySet.end()) {mySet.erase(s[left]);left++;}mySet.insert(s[right]);resLen = max(resLen, right - left + 1);right++;}return resLen; *//* 动态规划 */int size = s.size();if (size == 0 || size == 1) {return size;}map<char, int> myMap;vector<int> dp(size, -1);dp[0] = 1;for(int i = 0; i < size; ++i) {myMap[s[i]] = -1;}myMap[s[0]] = 0;for(int i = 1; i < size; ++i) {if (i - myMap[s[i]] > dp[i - 1]) {dp[i] = dp[i - 1] + 1;} else {dp[i] = i - myMap[s[i]];}myMap[s[i]] = i;}int max = -1;for(int i = 0; i < size; ++i) {if (dp[i] > max) {max = dp[i];}}return max;}
};

23 剑指 Offer 54. 二叉搜索树的第k大节点力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:void dfs(TreeNode* root, vector<int>& vec) {if (!root) {return;}dfs(root -> left, vec);vec.push_back(root -> val);dfs(root -> right, vec);}int kthLargest(TreeNode* root, int k) {if (!root) {return 0;}vector<int> vec;dfs(root, vec);int index = vec.size() - k;return vec[index];}
};

24 剑指 Offer 61. 扑克牌中的顺子力扣

class Solution {
public:bool isStraight(vector<int>& nums) {sort(nums.begin(), nums.end());int count = 0;int wang = 0;for(int i = 0; i < 4; ++i) {if (nums[i] == 0) {wang++;}if (nums[i] == nums[i+ 1] && nums[i] != 0) {return false;} if (nums[i + 1] > nums[i] && nums[i] != 0) {count += (nums[i + 1] - nums[i] - 1);}}if (wang >= count) {return true;}return false;}
};

25 剑指 Offer 63. 股票的最大利润剑指 Offer 63. 股票的最大利润 - 力扣（LeetCode） (leetcode-cn.com)

class Solution {
public:int maxProfit(vector<int>& prices) {//如果prices[i]大于min，则去更新一下利润res//否则说明当前的prices[i]比min还小，则更新minint size = prices.size();if (size == 0) {return 0;}int min = prices[0];int optimum = 0;for (int i = 1; i < size; ++i) {if (prices[i] <= min) {min = prices[i];} else if (prices[i] > min) {optimum = max(prices[i] - min, optimum);}}return optimum;}
};

26 剑指 Offer 32 - III. 从上到下打印二叉树 III 力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> res;if (!root) {return res;}queue<TreeNode*> qu;vector<int> tmpVec;qu.push(root);bool flag = false;while(!qu.empty()) {int size = qu.size();for (int i = 0; i < size; ++i) {TreeNode* tmpNode = qu.front();qu.pop();tmpVec.push_back(tmpNode -> val);if (tmpNode -> left) {qu.push(tmpNode -> left);}if (tmpNode -> right) {qu.push(tmpNode -> right);}}flag = !flag;if (flag == true) {res.push_back(tmpVec);} else {reverse(tmpVec.begin(), tmpVec.end());res.push_back(tmpVec);}tmpVec.clear();}return res;}
};

27 剑指 Offer 32 - II. 从上到下打印二叉树 II 力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> res;if (!root) {return res;}queue<TreeNode*> qu;vector<int> tmpVec;qu.push(root);bool flag = false;while(!qu.empty()) {int size = qu.size();for (int i = 0; i < size; ++i) {TreeNode* tmpNode = qu.front();qu.pop();tmpVec.push_back(tmpNode -> val);if (tmpNode -> left) {qu.push(tmpNode -> left);}if (tmpNode -> right) {qu.push(tmpNode -> right);}}flag = !flag;if (flag == true) {res.push_back(tmpVec);} else {reverse(tmpVec.begin(), tmpVec.end());res.push_back(tmpVec);}tmpVec.clear();}return res;}
};

28 剑指 Offer 56 - I. 数组中数字出现的次数力扣

看下这两个帖子基础知识 https://blog.csdn.net/qq_19018277/article/details/103812534?utm_medium=distribute.pc_relevant.none-task-blog-2%7Edefault%7EBlogCommendFromMachineLearnPai2%7Edefault-4.essearch_pc_relevant&depth_1-utm_source=distribute.pc_relevant.none-task-blog-2%7Edefault%7EBlogCommendFromMachineLearnPai2%7Edefault-4.essearch_pc_relevant

leetcode算法总结 —— 位运算_Charles Ren's Tech Blog-CSDN博客

class Solution {
public:vector<int> singleNumbers(vector<int>& nums) {int two = 2;int size = nums.size();vector<int> part1;vector<int> part2;vector<int> res;int  x = 0;for (int i = 0; i < size; ++i) {x = x ^ nums[i];}//从右向左找第一个1，记录这个1的位置，说明两个单数该位不同，再依据这位不同将数组分成两组，则两个单数一定分别在两组，两组各自抑或，找到各自的单数// 获取一个数二进制最低位的1int y = x & (-x);for (int i = 0; i < size; ++i) {int w = nums[i] & y;if (w == y) {part1.push_back(nums[i]);} else {part2.push_back(nums[i]);}}int z = 0;for (int i = 0; i < part1.size(); ++i) {z= z^ part1[i];}res.push_back(z);int m = 0;for (int i = 0; i < part2.size(); ++i) {m= m^ part2[i];}res.push_back(m);return res;}
};

29 剑指 Offer II 073. 狒狒吃香蕉力扣

class Solution {
public:int times(int index, int speed, vector<int>& piles) {if (piles[index] % speed != 0) {return piles[index] / speed + 1; } else {return piles[index] / speed;}}int minEatingSpeed(vector<int>& piles, int h) {sort(piles.begin(), piles.end());int size = piles.size();int right = piles[size - 1];double sum = 0;int left = 1;int res = right;while (left <= right) {int mid = (left + right) / 2;int tmpsum = 0;for (int i = 0; i < size; ++i) {tmpsum += times(i, mid, piles);}if (tmpsum > h) {left = mid + 1;} else {if (mid < res) {res = mid;}right = mid - 1;}}return res;}
};

30 剑指 Offer 56 - II. 数组中数字出现的次数 II 力扣

class Solution {
public:int singleNumber(vector<int>& nums) {/* 如果把所有的出现三次的数字的二进制表示的每一位都分别加起来，那么每一位都能被3整除。 我们把数组中所有的数字的二进制表示的每一位都加起来。如果某一位能被3整除，那么这一位对只出现一次的那个数的这一肯定为0。如果某一位不能被3整除，那么只出现一次的那个数字的该位置一定为1. */vector<int> vec(32, 0);int sum = 0;int size = nums.size(); for (int i = 0; i < size; ++i) {for (int j = 0; j < 32; ++j) {vec[j] += nums[i] & 1;nums[i] = (nums[i] >> 1);}}for (int i = 0; i < 32; ++i) {if (vec[i] % 3 == 1) {sum += pow(2, i);}}return sum;}
}

31 剑指 Offer 57 - II. 和为s的连续正数序列力扣

class Solution {
public:vector<vector<int>> findContinuousSequence(int target) {vector<vector<int>> res;if (target == 1) {return res;}int left = 1, right = left + 1;vector<int> tmpvec;while (right <= target / 2 + 1 && left <= right) {int tmpSum = 0;for (int i = left; i <= right; i ++) {tmpSum += i;}if (tmpSum < target) {right++;tmpSum += right;}else if (tmpSum > target) {left++;tmpSum -= left;}else if (tmpSum == target) {for (int m = left; m <= right; ++m) {tmpvec.push_back(m);}res.push_back(tmpvec);tmpvec.clear();left++;right++;}}return res;}
};

32 剑指 Offer 58 - I. 翻转单词顺序力扣

class Solution {
public:string reverseWords(string s) {stack<string> sta;int len = s.length();if (len == 0) {return s;}int i = 0;string tmpstr = "";while (i < len) {while (i < len && s[i] == ' ') {++i;}while (i < len && s[i] != ' ') {tmpstr += s[i];++i;}sta.push(tmpstr);tmpstr = "";}string res;while(!sta.empty()) {res += sta.top();sta.pop();res += " ";int l = res.size();if (sta.empty()) {if (res[l - 1] == ' ') {res = res.substr(0, l - 1);}}}int j = 0;int x = res.size();while (j < x && res[j] == ' ') {j++;}return res.substr(j, x - j);}
};

33 剑指 Offer 59 - II. 队列的最大值力扣

class MaxQueue {
public:queue<int> qu;deque<int> deq;MaxQueue() {}int max_value() {if (qu.empty() || deq.empty()) {return -1;}return deq.front();}void push_back(int value) {qu.push(value);while (!deq.empty() && value > deq.back()) {deq.pop_back();}deq.push_back(value);}int pop_front() {if (qu.empty()) {return -1;}int tmp = qu.front();qu.pop();if (!deq.empty() && deq.front() == tmp) {deq.pop_front();}return tmp;}
};

34 剑指 Offer II 002. 二进制加法力扣

class Solution {
public:string addBinary(string a, string b) {int lena = a.size();int lenb = b.size();int carry = 0;string res = "";if (lena > lenb) {string tmp = string(lena - lenb, '0');tmp += b;b = tmp;} else if (lena < lenb) {string tmp = string(lenb - lena, '0');tmp += a;a = tmp;}int cur = b.size() - 1;while (cur >= 0) {if ((a[cur] - '0') + (b[cur] - '0') + carry == 0) {res += "0";carry = 0;}else if ((a[cur] - '0') + (b[cur] - '0') + carry == 1) {res += "1";carry = 0;}else if ((a[cur] - '0') + (b[cur] - '0') + carry == 2) {res += "0";carry = 1;}else if ((a[cur] - '0') + (b[cur] - '0') + carry == 3) {res += "1";carry = 1;}cur--;}if (carry == 1) {res += "1";}reverse(res.begin(), res.end());return res;}
};

35 剑指 Offer II 003. 前 n 个数字二进制中 1 的个数

力扣

class Solution {
public:vector<int> countBits(int n) {/* 动态规划: 使用dp[i]记录i中1的个数； 如果i是奇数，二进制1的个数等于dp[i-1] + 1; 如果i是偶数，相当于(i / 2)左移一位，1的个数和i/2相等。 */vector<int> dp(n + 1, 0);dp[0] = 0;for (int i = 1; i <= n; ++i) {if (i % 2 == 1) {dp[i] = dp[i - 1] + 1;} else {dp[i] = dp[i / 2];}}return dp;}
};

36 剑指 Offer II 005. 单词长度的最大乘积

class Solution {
public:int maxProduct(vector<string>& words) {int size = words.size();vector<int> tmp(26, 0);//num为向量行数，0为初始化值vector<vector<int>> vec(size, tmp);for (int i = 0; i < size; ++i) {for (int j = 0; j < words[i].size(); ++j) {vec[i][words[i][j] - 'a']++;}}int ret = 0;int k, i, j;for (i = 0; i < size; ++i) {for (j = i + 1; j < size; ++j) {for (k = 0; k < 26; ++k) {if (vec[i][k] != 0 && vec[j][k] != 0) {break;}}if (k == 26) {if (words[i].size() * words[j].size() > ret) {ret = words[i].size() * words[j].size();}}}}return ret;}
};

37 剑指 Offer II 007. 数组中和为 0 的三个数力扣

class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {int len = nums.size();set<vector<int>> res;sort(nums.begin(), nums.end());int i = 0;for(; i < len-2; i++) {int j = i;if (nums[i] > 0) {break;}while(nums[i] == nums[i+1] && i < len-2) i++;int num = nums[j];int target = -num;int l = j+1, r = len-1;while(l < r) {int temp = nums[l] + nums[r];if(temp == target) {res.insert({num, nums[l], nums[r]});while(l + 1 < len && nums[l] == nums[l + 1]) {l++;}while(r > l && nums[r] == nums[r-1]) {r--;}l++,r--;} else if(temp > target) {r--;} else {l++;}}}vector<vector<int>> ans;ans.assign(res.begin(), res.end());return ans;}
};

38 剑指 Offer II 009. 乘积小于 K 的子数组

力扣

class Solution {
public:int numSubarrayProductLessThanK(vector<int>& nums, int k) {// 当 k = 0 直接返回 0 即可if(k == 0) return 0;int count = 0; // 记录乘积小于 k 的连续的子数组的个数。int product = 1; // 当前窗口内的数的乘积int left = 0; // 窗口左边界for(int right = 0; right < nums.size(); right++){product *= nums[right];// 当 窗口内数的乘积 >= k ，将窗口左边界向右移，同时更新窗口内数乘积while (left <= right && product >= k){product /= nums[left++];}// 统计该窗口中乘积小于 k 的连续的子数组的个数。count += (right - left + 1);}return count;}
};

39 剑指 Offer II 010. 和为 k 的子数组

力扣

class Solution {
public:int subarraySum(vector<int>& nums, int k) {/* vector<int> sum(nums.size() + 1);for (int i = 0; i < nums.size(); ++i) {sum[i + 1] = sum[i] + nums[i];}int ret = 0;for (int i = 0; i < nums.size(); ++i) {for (int j = 0; j <= i; ++j) {if (sum[i + 1] - sum[j] == k) {ret++;}}}return ret; *///前缀和+hashmap组合//用于存前缀和的个数unordered_map<int, int> mp;int sum = 0;int res = 0;mp[0] = 1;for(int i=0; i<nums.size();i++) {sum += nums[i];if(mp.count(sum-k) !=0) {res += mp[sum-k];}if(mp.count(sum) == 0) {mp[sum] = 1;}else {mp[sum]++;}}return res;}
};

40 剑指 Offer II 011. 0 和 1 个数相同的子数组

力扣

class Solution {
public:/* 剑指 Offer II 010. 和为 k 的子数组  和这道题做法一样，如果这道题把0看成-1 做法就完全一样 */int findMaxLength(vector<int>& nums) {int size = nums.size();unordered_map<int, int> mymap;int sum = 0;int ret = 0;int x = -1;for (int i = 0; i < size; ++i) {if (nums[i] == 0) {sum += x;} else {sum += 1;}if (sum == 0 && ret < i + 1) {ret = i + 1;}if (mymap.find(sum) == mymap.end()) {mymap[sum] = i;} else {if (ret < i - mymap[sum]) {ret = i - mymap[sum];}}}return ret;}
};

41 剑指 Offer II 013. 二维子矩阵的和力扣

class NumMatrix {
public:vector<vector<int>> vec;NumMatrix(vector<vector<int>>& matrix) {int rows = matrix.size();int cols = matrix[0].size();vec.resize(rows + 1, vector<int>(cols + 1, 0)); // rows和cols都加一是为了在sumRegion里少if，否则可以看下注释里if很多for (int i = 0; i < rows; ++i) {for (int j = 0; j < cols; ++j) {vec[i + 1][j + 1] = vec[i + 1][j] - vec[i][j] + matrix[i][j] + vec[i][j + 1];}}}int sumRegion(int row1, int col1, int row2, int col2) {//if (row1 > 0 && col1 > 0) {return vec[row2 + 1][col2 + 1] - vec[row1][col2 + 1] - vec[row2 + 1][col1] + vec[row1][col1]; //return vec[row2][col2] - vec[row1 - 1][col2] - vec[row2][col1 - 1]  + vec[row1 - 1][col1 - 1]; /* } else if (row1 == 0 && col1 == 0) {return vec[0][0];  }else if (col1 == 0) {return vec[row2][col2] - vec[row1-1][col2];} else if (row1 == 0) {return vec[row2][col2] - vec[row2][col1 - 1];} return -1; */}
};

42 剑指 Offer II 012. 左右两边子数组的和相等

力扣

class Solution {
public:int pivotIndex(vector<int>& nums) {int len = nums.size();vector<int> sums(len+1);for(int i = 1;i <= len;++i){sums[i] = sums[i-1] + nums[i-1];}for(int i = 0;i < len;++i){if(sums[i] == (sums[len] - sums[i+1])) return i;}return -1;}
};

43 剑指 Offer II 041. 滑动窗口的平均值

力扣

class MovingAverage {
public:/** Initialize your data structure here. */double len;deque<int> deq;double sum;MovingAverage(int size) {len = size;}double next(int val) {if (deq.size() < len) {deq.push_back(val);sum += val;return sum / deq.size();} else {int top = deq.front();deq.pop_front();sum -= top;deq.push_back(val);sum += val;return sum / deq.size();}return 0.0;}
};/*** Your MovingAverage object will be instantiated and called as such:* MovingAverage* obj = new MovingAverage(size);* double param_1 = obj->next(val);*/

44 剑指 Offer II 016. 不含重复字符的最长子字符串力扣

class Solution {
public:int lengthOfLongestSubstring(string s) {int size = s.size();int left = 0;int right = 0;unordered_map<char, int> mp;int len = 0;int ret = 0;for (int i = 0; i < size; ++i) {mp[s[i]]++;while (left <= i && mp[s[i]] > 1) {mp[s[left]]--;left++;}ret = max (ret, i - left + 1);}return ret;}
};

45 剑指 Offer II 018. 有效的回文力扣剑指 Offer II 018. 有效的回文

class Solution {
public:bool isPalindrome(string s) {int i = 0;int j = s.size() - 1;while (i < j) {if (!isalnum(s[i])) {i++;}else if (!isalnum(s[j])) {j--;}else {if (tolower(s[i]) != tolower(s[j])) {return false;}else {i++;j--;}} }return true;}
};/* isalpha ：判断一个字符是否为字母，如果是则返回true，否则返回false；
isdigit : 判断一个字符是否表示数字，如果是则返回true，否则返回false；
isalnum : 判断一个字符是否表示数字或者字母，如果是则返回true，否则返回false;
islower : 判断一个字符是否为小写字母，如果是则返回true，否则返回false;
isupper : 判断一个字符是否为大写字母，如果是则返回true，否则返回false；
tolower : 若字符为字母则转化为小写字母；
toupper : 若字符为字母则转化为大写字母；

46 剑指 Offer II 021. 删除链表的倒数第 n 个结点

力扣

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode* dummy = new ListNode(0, head);//引入虚假头结点真的就是非常好用ListNode* fast = head;ListNode* slow = dummy;for (int i = 0; i < n; ++i) {fast = fast->next;}while (fast) {fast = fast->next;slow = slow->next;}slow->next = slow->next->next;ListNode* ans = dummy->next;delete dummy;return ans;}
};

47剑指 Offer II 022. 链表中环的入口节点

力扣


/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode *detectCycle(ListNode *head) {/* // 可用map统计，看某一个节点是否被指向过if(head==NULL) return NULL;unordered_map<ListNode*, int> my_map;ListNode* tmp = head;while(tmp!=NULL) {if(my_map.find(tmp)==my_map.end()) {my_map[tmp] = 1;tmp = tmp->next;}else {return tmp;}}return NULL; *//* 假设链表全长为 m+n，其中 m 表示环入口之前的长度，n 表示环的长度 当两指针第一次相遇，这时候快指针一定已经饶了环一圈了，我们假设相遇的位置距离环入口k，此时有下列状态： distance(fast) = m + n + k distance(slow) = m + k m + n + k = 2m + 2k ==> n = m + k ==> m = n - k 然后我们将快指针设回头节点，两根指针这时候同时按照步长为1前进，那么当慢指针走到环入口时，有以下状态： distance(fast) = distance(slow) = n - k 不难发现此时快指针也刚好走到环入口 *//*class Solution {
public:ListNode *detectCycle(ListNode *head) {ListNode *fast = head, *slow = head;while (true) {if (fast == nullptr || fast->next == nullptr) return nullptr;fast = fast->next->next;slow = slow->next;if (fast == slow) break;}fast = head;while (slow != fast) {slow = slow->next;fast = fast->next;}return fast;}
};作者：jyd
链接：https://leetcode-cn.com/problems/c32eOV/solution/jian-zhi-offer-ii-022-lian-biao-zhong-hu-8f1m/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
*/}
};

48 剑指 Offer II 024. 反转链表

力扣

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {/* if (head == nullptr|| head -> next == nullptr) {return head;}stack<ListNode*> st;ListNode* cur = head;while(cur) {st.push(cur);cur = cur -> next;}ListNode* ret = st.top();ListNode* res = st.top();st.pop();while (!st.empty()) {ret -> next = st.top();ret = ret -> next;st.pop();}ret -> next = nullptr;return res; */// 调用递推公式反转当前结点之后的所有节点// 返回的结果是反转后的链表的头结点/* if (head == nullptr || head -> next == nullptr) {return head;}ListNode* newHead = reverseList(head->next);head->next->next = head;head->next = nullptr;return newHead;*/if (head == nullptr || head -> next == nullptr) {return head;}ListNode* pre = nullptr;ListNode* now = head;ListNode* after;while (now) {after = now -> next;now -> next = pre;pre = now;now = after; }return pre;}
};

49 剑指 Offer II 026. 重排链表

力扣

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseLi(ListNode* head) {if (head == nullptr || head -> next == nullptr) {return head;}ListNode* pre = nullptr;ListNode* cur = head;ListNode* tmp_next = nullptr;while(cur) {tmp_next = cur -> next;cur -> next = pre;pre = cur;cur = tmp_next;}return pre; }void reorderList(ListNode* head) {int lenhead = 0;ListNode* cur1 = head;while (cur1) {cur1 = cur1 -> next;lenhead++;}int tag;if (lenhead % 2== 1) {tag = lenhead / 2 + 1;} else{tag = lenhead / 2;}ListNode* taga = head;while (tag > 0) {taga =taga-> next;tag--;}ListNode* pre = head;while (pre -> next != taga) {pre = pre -> next;}pre -> next = nullptr;ListNode* rever = reverseLi(taga); ListNode* h1 = head;ListNode* h2 = rever;while (h1 && h2) {ListNode* tmp1 = h1 -> next;ListNode* tmp2 = h2 -> next;h1 -> next = h2;h2 -> next= tmp1;h1 = tmp1;h2 = tmp2;}}
};

50 剑指 Offer II 027. 回文链表

力扣

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {/* if (head == nullptr || head -> next == nullptr) {return head;}ListNode* pre = nullptr;ListNode* cur = head;ListNode* tmp_next = nullptr;while(cur) {tmp_next = cur -> next;cur -> next = pre;pre = cur;cur = tmp_next;}return pre; */if (head == nullptr || head -> next == nullptr) {return head;}ListNode* tmp = reverseList(head -> next);head -> next -> next = head;head -> next = nullptr;return tmp;}bool isPalindrome(ListNode* head) {if (head -> next == nullptr || head == nullptr) {return true;}int len = 0;ListNode* cur = head;while(cur) {len++;cur = cur -> next;}ListNode* cur2 = head;int dis = len / 2;if (len % 2 == 1) {while (dis >= 0) {cur2 = cur2 -> next;dis--;}} else {while (dis >= 1){cur2 = cur2 -> next;dis--;}}ListNode* cur3 = reverseList(cur2);ListNode* cur5 = head;int i = 1;while (i <= len / 2) {if (cur3 -> val != cur5 -> val) {return false;}cur5 = cur5 -> next;cur3 = cur3 -> next;i++;}return true;}
};

51 剑指 Offer II 028. 展平多级双向链表

力扣

/*
// Definition for a Node.
class Node {
public:int val;Node* prev;Node* next;Node* child;
};
*/
class Solution {
public:vector<Node*> vec;void dfs(Node* head) {if (!head) {return;}vec.push_back(head);dfs(head -> child);dfs(head -> next);}Node* flatten(Node* head) {if (!head) {return head;}dfs(head);Node* pree = nullptr;for (int i = 0; i < vec.size() - 1; ++i) {cout << vec[i] -> val;vec[i + 1]->prev = vec[i];vec[i] -> next = vec[i + 1];vec[i] -> child  =  nullptr;}return vec[0];}
};

52 剑指 Offer II 030. 插入、删除和随机访问都是 O(1) 的容器

力扣

生产随机数如何在C++中产生随机数 - Eventide - 博客园

class RandomizedSet {
public:map<int, int> mymap;vector<int> vec;RandomizedSet() {}bool insert(int val) {if (mymap.count(val) != 0) {return false;}vec.push_back(val);mymap[val] = vec.size() - 1;return true;}bool remove(int val) {if (mymap.count(val) == 0) {return false;}int loc = mymap[val];vec[loc] = vec[vec.size() - 1];mymap[vec[vec.size() - 1]] = loc;vec.pop_back();mymap.erase(val);return true;}int getRandom() {return vec[rand() % vec.size()];}
};
/*** Your RandomizedSet object will be instantiated and called as such:* RandomizedSet* obj = new RandomizedSet();* bool param_1 = obj->insert(val);* bool param_2 = obj->remove(val);* int param_3 = obj->getRandom();*/

53 剑指 Offer II 076. 数组中的第 k 大的数字剑指 Offer II 076. 数组中的第 k 大的数字 - 力扣（LeetCode） (leetcode-cn.com)

class Solution {
public:// less<int>  top是最大// greater<int>  top是最大// pair以及优先队列中用pair https://blog.csdn.net/hnjzsyjyj/article/details/108929993int findKthLargest(vector<int>& nums, int k) {priority_queue<int, vector<int>, greater<int>> vec; // top是最小的int size = nums.size();for (int i = 0; i < size; ++i) {if (i < k) {vec.push(nums[i]);} else {if (vec.top() < nums[i]) {vec.pop();vec.push(nums[i]);}}}return vec.top();}
};/*vector中的元素是pair，如何对vector进行排序，排序原则是pair的first或者second#include<cmath>
#include<algorithm>
#include <vector>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
pair<int, int> p;
vector<pair<int, int>> vec;
int cmp(pair<int, int>a, pair<int, int>b){if (a.first != b.first)return a.first>b.first;else return a.second<b.second;
}
int main(){int a, b;p = make_pair(1, 2);vec.push_back(p);p = make_pair(3, 4);vec.push_back(p);p = make_pair(5, 6);vec.push_back(p);sort(vec.begin(), vec.end(), cmp);cout << vec.size() << endl;for (int i = 0; i < vec.size(); ++i) {cout << vec[i].first << " " << vec[i].second << endl;}return 0;
}
*//* #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef struct example
{int elem1;int elem2;
}example;/*这个comparison函数很重要.如果希望升序排序,就是"<",降序排列就是">"号,这样便于直观记忆.如果希望用elem2作为比较标准
就把elem1改为elem2,这样结构体就以elem2为比较标准排序了.*/
bool comparison(example a,example b){return a.elem1<b.elem1;
}
int main()
{int N;fin>>N;vector<example> array(N);for(int i=0;i<N;i++){fin>>array[i].elem1>>array[i].elem2;}sort(array.begin(),array.end(),comparison);for(int i=0;i<N;i++){cout<<array[i].elem1<<" "<<array[i].elem2<<endl;}return 0;*/

54 剑指 Offer II 033. 变位词组

力扣

class Solution {
public:vector<vector<string>> groupAnagrams(vector<string>& strs) {// 建立一个map，str排序后做key，<vector<string>做value，在map中能找到的key就放入mapvector<string> vec;map<string, vector<string>> mymap;int size = strs.size();vector<vector<string>> res;for (int i = 0; i < size; ++i) {string tmp = strs[i];sort(strs[i].begin(), strs[i].end());mymap[strs[i]].push_back(tmp);}map<string, vector<string>>::iterator iter = mymap.begin();for (; iter != mymap.end(); ++iter) {res.push_back(iter -> second);}return res;}
};

55 剑指 Offer II 068. 查找插入位置

力扣

class Solution {
public:int searchInsert(vector<int>& nums, int target) {int left = 0, right = nums.size() - 1;while (left <= right) {int mid = left + ((right - left) >> 1);if (nums[mid] == target) {return mid;} else if (nums[mid] > target) {right = mid - 1;} else {left = mid + 1;}}return left;}
};

56 剑指 Offer II 081. 允许重复选择元素的组合

力扣

class Solution {
public:vector<int> tmpVec;vector<vector<int>> res;int i = 0;void dfs(int i, int target, int sum, vector<int>& candidates) {if (sum == target) {res.push_back(tmpVec);}if (sum < target) {for (; i < candidates.size(); ++i) {tmpVec.push_back(candidates[i]);sum += candidates[i];dfs(i, target, sum, candidates);tmpVec.pop_back();sum -= candidates[i];}}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {dfs(0, target, 0, candidates);return res;}
};

57 剑指 Offer II 082. 含有重复元素集合的组合力扣

class Solution {
public:vector<int> tmpVec;vector<vector<int>> res;int i = 0;set<vector<int>> myset;void dfs(int i, int target, int sum, vector<int>& candidates) {if (sum == target) {res.push_back(tmpVec);}if (sum < target) {for (; i < candidates.size(); ++i) {tmpVec.push_back(candidates[i]);sum += candidates[i];dfs(i + 1, target, sum, candidates);tmpVec.pop_back();sum -= candidates[i];}}}vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {dfs(0, target, 0, candidates);for (int i = 0; i < res.size(); ++i) {sort(res[i].begin(), res[i].end());myset.insert(res[i]);}vector<vector<int>> res2;set<vector<int>, int>::iterator iter = myset.begin();for (; iter != myset.end(); ++iter) {res2.push_back(*iter);}return res2;}
};

58 剑指 Offer II 083. 没有重复元素集合的全排列

力扣

class Solution {
public:vector<vector<int>> res;vector<int> tmp;void dfs(vector<int>& nums, vector<int>& visit) {if (tmp.size() == nums.size()) {res.push_back(tmp);//tmp.clear();return;}for (int i = 0; i < nums.size(); ++i) {if (visit[i] == 0) {tmp.push_back(nums[i]);visit[i] = 1;dfs(nums, visit);visit[i] = 0;tmp.pop_back();}}}vector<vector<int>> permute(vector<int>& nums) {int size = nums.size();vector<int> visit(size, 0);dfs(nums, visit);return res;}
};

59 剑指 Offer II 069. 山峰数组的顶部

力扣

class Solution {
public:int peakIndexInMountainArray(vector<int>& arr) {int size = arr.size();int left = 0;int right = size - 1;int mid;int res;while (left <= right) {mid = (left + right) / 2;if (arr[mid] < arr[mid + 1]) {left = mid + 1;}if (arr[mid] > arr[mid + 1]) {res = mid;right = mid - 1;}}return res;}
};

60 剑指 Offer II 035. 最小时间差

力扣

class Solution {
public:int StrToTIme(string str) {string hour = str.substr(0, 2);string minute = str.substr(3, 2);int h = (hour[0] - '0') * 10 + (hour[1] - '0');int m = (minute[0] - '0') * 10 + (minute[1] - '0');return h * 60 + m;}int findMinDifference(vector<string>& timePoints) {vector<int> minGrp;int size = timePoints.size();for (int i = 0; i < size; ++i) {minGrp.push_back(StrToTIme(timePoints[i]));}sort(minGrp.begin(), minGrp.end());int ret = INT_MAX, Size = minGrp.size();for (int i = 0; i < Size - 1; ++i) {if (minGrp[i + 1] - minGrp[i] < ret) {ret = minGrp[i + 1] - minGrp[i];}}if (1440 + minGrp[0] - minGrp[Size - 1] < ret) {ret = 1440 + minGrp[0] - minGrp[Size - 1];}return ret;}
};

61 剑指 Offer II 038. 每日温度

力扣

class Solution {
public:vector<int> dailyTemperatures(vector<int>& temperatures) {// 单调栈// 第一个元素入栈，若下一个元素比栈顶元素大  则找到了比栈顶元素大的第一个元素，弹出栈顶元素并记录结果值，不断反复弹出  直到找到比当前元素大的栈顶元素，再把当前元素放入栈// 栈中存的是下标int size = temperatures.size();vector<int> res(size, 0);stack<int> sta;for (int i = 0; i < size; ++i) {if (i == 0) {sta.push(0);}while(sta.size() != 0 && temperatures[i] > temperatures[sta.top()]) {res[sta.top()] = i - sta.top();sta.pop();}sta.push(i);}return res;}
};

62 剑指 Offer II 036. 后缀表达式

力扣

class Solution {
public:int evalRPN(vector<string>& tokens) {/* 栈中只保存操作数，操作符不需要保存在栈中
如果遇到的是一个操作数，则将其入栈；如果遇到的是一个操作符，则两个操作数出栈并执行相应的运算，然后计算结果入栈 */stack<int> sta;int size = tokens.size();for (int i = 0; i < size; ++i) {if (tokens[i] != "+" && tokens[i] != "-" &&  tokens[i] != "*" && tokens[i] != "/") {sta.push(stoi(tokens[i]));} else {int right = sta.top();sta.pop();int left = sta.top();sta.pop();int ret = 0;if (tokens[i] == "+") {ret = left + right;}if (tokens[i] == "-") {ret = left - right;}if (tokens[i] == "*") {ret = left * right;}if (tokens[i] == "/") {ret = left / right;cout << ret;}/* switch (tokens[i]) {case "+":ret = left + right;break;case "-":ret = left - right;break;case "*":ret = left * right;break;default:ret = left / right;} */sta.push(ret);}}return sta.top();}
};

63 剑指 Offer II 037. 小行星碰撞

力扣

class Solution {
public:vector<int> asteroidCollision(vector<int>& asteroids) {stack<int> sta;int size = asteroids.size();for (int i = 0; i < size; ++i) {while (sta.size() > 0 && sta.top() > 0 && asteroids[i] < 0 && sta.top() < (-asteroids[i])) {sta.pop();}if (sta.size() > 0 && sta.top() > 0 && asteroids[i] < 0 && sta.top() == (-asteroids[i])) {sta.pop();}else if (sta.size() == 0 || asteroids[i] > 0 || sta.top() < 0) {sta.push(asteroids[i]);}}vector<int> res;while (sta.size() != 0) {res.push_back(sta.top());sta.pop();}reverse(res.begin(), res.end());return res;}
};

64 剑指 Offer II 049. 从根节点到叶节点的路径数字之和

力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/class Solution {
public:int dfs(TreeNode* root, int path) {if (!root) {return 0;}path = path * 10 + root -> val;if (root -> left == nullptr && root -> right == nullptr) {return path;}return dfs(root -> left, path) + dfs(root -> right, path);}int sumNumbers(TreeNode* root) {return dfs(root, 0);}
};

65 剑指 Offer II 105. 岛屿的最大面积力扣

class Solution {
public:int dfs(vector<vector<int>>& grid, int rows, int cols, int row, int col) {if (row < 0 || col < 0 || row >= rows || col >= cols || grid[row][col] == 0) {return 0;}grid[row][col] = 0;return 1 + dfs(grid, rows, cols, row + 1, col) + dfs(grid, rows, cols, row, col + 1) + dfs(grid, rows, cols, row - 1, col) + dfs(grid, rows, cols, row, col - 1);}int maxAreaOfIsland(vector<vector<int>>& grid) {int rows = grid.size();int cols = grid[0].size();int tmpMax = -1;for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; ++j) {int t = dfs(grid, rows, cols, i, j);if (tmpMax < t) {tmpMax = t;}}}return tmpMax;}
};

66 剑指 Offer II 058. 日程表

力扣

class MyCalendar {
public:vector<vector<int> > v = vector<vector<int> > (1000, vector<int>(2, 0));int top = 0;MyCalendar() {}bool book(int start, int end) {if (top == 0) {v[0][0] = start;v[0][1] = end;top = 1;return true;}for (int i = 0; i < top; ++i) {if (!(start >= v[i][1] || end <= v[i][0])) {return false;}}v[top][0] = start;v[top][1] = end;top++;return true;}
};/*** Your MyCalendar object will be instantiated and called as such:* MyCalendar* obj = new MyCalendar();* bool param_1 = obj->book(start,end);*/

67 剑指 Offer II 045. 二叉树最底层最左边的值

力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:vector<vector<int>> res;vector<int> v;queue<TreeNode*> qu;int findBottomLeftValue(TreeNode* root) {qu.push(root);while(!qu.empty()) {int size = qu.size();for (int i = 0; i < size; ++i) {TreeNode* tmp = qu.front();v.push_back(tmp -> val);qu.pop();if (tmp -> left) {qu.push(tmp -> left);}if (tmp -> right) {qu.push(tmp -> right);}}res.push_back(v);v.clear();}int layers = res.size();return res[layers - 1][0];}
};

68 剑指 Offer II 089. 房屋偷盗

力扣

class Solution {
public:int rob(vector<int>& nums) {int size = nums.size();vector<int> vec(size + 1, 0);vector<int> visit(size + 1, 0);if (size == 1) {return nums[0];}vec[0] = nums[0];vec[1] = max(nums[0], nums[1]);int sum = 0;for (int i = 2; i < size; ++i) {vec[i] = max(vec[i - 1], vec[i - 2] + nums[i]);}return vec[size -1];}
};
/* https://leetcode-cn.com/problems/Gu0c2T/solution/jian-zhi-offer-2-mian-shi-ti-89-shu-zhon-86xo/ */

69 剑指 Offer II 099. 最小路径之和

力扣

class Solution {
public:int minPathSum(vector<vector<int>>& grid) {int size = grid.size();vector<vector<int>> vec(size, vector<int>(grid[0].size(), 0));vec[0][0] = grid[0][0];for (int i = 1; i< grid[0].size(); ++i) {vec[0][i] = vec[0][i - 1] + grid[0][i];}for (int i = 1; i< grid.size(); ++i) {vec[i][0] = vec[i - 1][0] + grid[i][0];}for (int i = 1; i < size; ++i) {for (int j = 1; j < grid[0].size(); ++j) {vec[i][j] = min(vec[i - 1][j], vec[i][j - 1]) + grid[i][j];}}return vec[size - 1][grid[0].size() - 1];}
};

70 剑指 Offer II 100. 三角形中最小路径之和

class Solution {
public:int minimumTotal(vector<vector<int>>& triangle) {int rows = triangle.size();int cols = triangle[rows - 1].size();vector<vector<int>> vec(rows, vector<int>(cols, 0)) ;vec[0][0] = triangle[0][0];for (int i = 1; i < rows; ++i) {vec[i][0] = vec[i - 1][0] + triangle[i][0];}for (int i = 1; i < rows; ++i) {for (int j = 1; j < i + 1; ++j) {vec[i][j] = min(vec[i - 1][j - 1], vec[i - 1][j])  + triangle[i][j];}vec[i][i] = vec[i - 1][i - 1] + triangle[i][i];}int res = INT_MAX;for (int i = 0; i < cols; ++i) {if (vec[rows - 1][i] < res) {res = vec[rows - 1][i];}}return res;}
};

71 剑指 Offer II 101. 分割等和子集

力扣

/*
设 f(i, j) 表示能否从前 i 个物品(物品编号为 0 ~ i - 1)中选择若干物品放满容量为 j 的背包。对于 f(i, j) 存在两个选择，第一个选择是将标号为 i - 1 的物品放入背包，如果能从前 i - 1 个物品中选择若干物品放满容量为 j - nums[i - 1] 的背包(即 f(i - 1, j - nums[i - 1]) 为 true)，那么 f(i, j) 为 true。另一个选择是不把标号为 i - 1 的物品放入背包，如果能从前 i - 1 个物品中选择若干物品放满容量为 j 的背包(即 f(i - 1, j) 为 true)，那么 f(i, j) 为 true。即当 j 等于 0 时，即背包容量为空，只要不选择物品就可以，所以 f(i, 0) 为 true。当 i 等于 0 时，即物品数量为 0，那么除了空背包都无法装满，所以当 j 大于 0 时，f(0, j) 为 fasle;作者：master_xue
链接：https://leetcode-cn.com/problems/NUPfPr/solution/jian-zhi-offer-2-mian-shi-ti-101-shu-zho-mt8e/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
*/class Solution {
public:bool canPartition(vector<int>& nums) {int sum = 0;for (auto& n : nums) {sum += n;}if (sum & 1 != 0) {return false;}int target = sum >> 1;vector<vector<bool>> dp(nums.size() + 1, vector<bool>(target + 1, false));dp[0][0] = true;for (int i = 1; i <= nums.size(); ++i) {for (int j = 0; j <= target; ++j) {dp[i][j] = dp[i - 1][j];if (!dp[i][j] && j >= nums[i - 1]) {dp[i][j] = dp[i - 1][j - nums[i - 1]];}}}return dp[nums.size()][target];}
};

72 剑指 Offer II 091. 粉刷房子

力扣

class Solution {
public:int minCost(vector<vector<int>>& costs) {int size = costs.size();vector<vector<int>> vec(100, vector<int>(3, 0));vec[0][0] = costs[0][0];vec[0][1] = costs[0][1];vec[0][2] = costs[0][2];for (int i = 1; i < size; ++i) {vec[i][0] = min(vec[i - 1][1], vec[i - 1][2]) + costs[i][0];vec[i][1] = min(vec[i - 1][0], vec[i - 1][2]) + costs[i][1];vec[i][2] = min(vec[i - 1][0], vec[i - 1][1]) + costs[i][2];}int res = 99999;for (int i = 0; i < 3; ++i) {if (vec[size - 1][i] < res) {res = vec[size - 1][i];}}return res;}
};
确定状态转换：
题目要求：相邻的两间房子不能涂相同的颜色，那只要我们当前要涂的颜色和前一间不同就好 （废话）
假设我们现在到了第3间房子，我们考虑一下涂那种颜色：
假设涂红色：所以前一家房子只能是蓝色或者绿色，从他们中选一个最便宜的吧！
那就是 dp[2][0] = Math.min(dp[1][1],dp[1][2])+costs[2][0]
蓝色和绿色同理，由此我们可以得到状态转换方程！dp[i][0] = Math.min(dp[i-1][1],dp[i-1][2])+costs[i][0];dp[i][1] = Math.min(dp[i-1][0],dp[i-1][2])+costs[i][1];dp[i][2] = Math.min(dp[i-1][0],dp[i-1][1])+costs[i][2];

73 剑指 Offer II 095. 最长公共子序列

力扣

class Solution {
public:int longestCommonSubsequence(string text1, string text2) {int size1 = text1.size();int size2 = text2.size();vector<vector<int>> vec(size1 + 1, vector<int> (size2 + 1, 0));// vec[i][j] = vec[i - 1][j - 1] + 1, text1[i] == text[j]// vec[i][j] - max(vec[ i - 1][j], vec[i][j - 1]), text1[i] != text[j]for (int i = 0; i < size1; ++i) {for (int j = 0; j < size2; ++j) {if (text1[i] == text2[j]) {vec[i + 1][j + 1] = vec[i][j] + 1;} else {vec[i + 1][j + 1] = max(vec[i][j + 1], vec[i + 1][j]);}}}return vec[size1][size2];}
};

74 剑指 Offer II 102. 加减的目标值

力扣

class Solution {
public:int res = 0;void dfs(vector<int>& nums, int index, int size, int sum, int target) {if (index == size) {if (sum == target) {res++;}return;}dfs(nums, index + 1, size, sum - nums[index], target);dfs(nums, index + 1, size, sum + nums[index], target);}int findTargetSumWays(vector<int>& nums, int target) {int size = nums.size();dfs(nums, 0, size, 0, target);return res;}
};

75 剑指 Offer II 098. 路径的数目

力扣

class Solution {
public:int uniquePaths(int m, int n) {int rows = m - 1;int cols = n - 1;vector<vector<int>> vec(m, vector<int>(n, 1));for (int i = 1; i < m; ++i) {for (int j = 1; j < n; ++j) {vec[i][j] = vec[i - 1][j] + vec[i][j - 1];} }return vec[m - 1][n - 1];}
};

76 剑指 Offer II 053. 二叉搜索树中的中序后继

力扣


/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:vector<TreeNode*> vec;void MidOrder(TreeNode* root) {if (!root) {return;}MidOrder(root -> left);vec.push_back(root);MidOrder(root -> right);}TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {MidOrder(root);int size = vec.size();for (int i = 0; i < size;  ++i) {if (i != size -1 && p == vec[i]) {return vec[i + 1];}}return nullptr;}
};

77 剑指 Offer II 075. 数组相对排序

力扣

class Solution {
public:vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {int size1 = arr1.size();vector<int> hash(size1, 0);int size2 = arr2.size();vector<int> res;for (int i = 0; i < size2; ++i) {for (int j = 0; j < size1; ++j) {if (arr2[i] == arr1[j] && hash[j] == 0) {res.push_back(arr1[j]);hash[j] = 1;}}}vector<int> tmpVec;for (int i = 0; i < size1; ++i) {if (hash[i] == 0) {tmpVec.push_back(arr1[i]);}}sort(tmpVec.begin(), tmpVec.end());for (int i = 0; i < tmpVec.size(); ++i) {res.push_back(tmpVec[i]);}return res;}
};

78 剑指 Offer II 052. 展平二叉搜索树力扣

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:vector<int> vec;void MidOrder(TreeNode* root) {if (!root) {return;}MidOrder(root -> left);vec.push_back(root -> val);MidOrder(root -> right);}TreeNode* increasingBST(TreeNode* root) {MidOrder(root);int size = vec.size();TreeNode* res = new TreeNode(0);TreeNode* tmp = res;for (int i = 0; i < size; ++i) {TreeNode* node = new TreeNode(vec[i]);node -> left = nullptr;tmp -> right = node;tmp = node;}return res -> right;}
};

79 剑指 Offer II 110. 所有路径

力扣

class Solution {
public:vector<vector<int>> res;vector<int> tmp;void dfs(vector<vector<int>>& graph, int size, int index) {if (index == size - 1) {res.push_back(tmp);}for (int i = 0; i < graph[index].size(); ++i) {tmp.push_back(graph[index][i]);dfs(graph, size, graph[index][i]);tmp.pop_back();}}vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {int size = graph.size();tmp.push_back(0);dfs(graph, size, 0);return res;}
};

80 剑指 Offer II 059. 数据流的第 K 大数值

力扣

class KthLargest {
public:int K;priority_queue<int, vector<int>, greater<int>> q; // top是最小的队列底是最大的，维持一个K大小的队列能把最大的K个数永远放对列vector<int> vec;KthLargest(int k, vector<int>& nums) {K = k;for (int i  = 0; i < nums.size(); ++i) {q.push(nums[i]);}}int add(int val) {q.push(val);while(q.size() > K) {q.pop();}return q.top();}
};/*** Your KthLargest object will be instantiated and called as such:* KthLargest* obj = new KthLargest(k, nums);* int param_1 = obj->add(val);*/

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