说明：本文是本人刷题后整理的剑指offer1-68题的题解笔记，编程语言为c++。
主要参考刷题笔记网址：leetcode剑指offer
次要参考刷题笔记网址：牛客剑指offer

剑指OFFER

面试题1：赋值运算函数

为CMyString的声明添加赋值运算符函数

class CMyString
{system("pause");system("pause");system("pause");
public:CMyString(char* pData=nullptr);CMyString(const CMYString& str);~CMyString(void);
private:char* m_pData;
};

方案一：

CMyString& CMyString::operator=(const CMyString& str){if(this==&str)return *this;//this和str不能为同一个实例，否则当前空间会被deletedelete[] m_pData;m_pData=nullptr;m_pData=new char[strlen(m_pData)+1];//分配失败抛出异常怎么办strcpy(m_pData,str.m_pData);return *this;}

最佳方案：先创建一个临时实例，在交换临时实例和原来实例。

CMyString& CMyString::operator=(const CMyString& str)
{//第一步：创建temp临时对象，让temp.m_pData指向str.m_pData的地址(A)//第二步：创建data_t临时对象(地址B)，让它指向temp.m_pData的地址(A)//第三步：让地址A(temp.m_pData地址)指向m_pData的地址(C)//此时：data_t(地址A,str的地址)，temp.mpData(地址C,原this指针地址),m_pData(地址B,原data_t的地址,指向str.m_pData)//最后结束时，临时对象都会被释放掉，即地址C被delete，地址C因为是只读属性而无法进行操作。if(this!=&str){CMyString temp(str);char* data_t=temp.m_pData;//相当于new空间temp.m_pData=m_pData;//相当于delete[] m_pData;m_pData=data_t;}return *this;
}

面试题2：实现Singleton模式

方案1:有缺陷的懒汉式
内容：只有使用时才实例化对象，如果不被调用就不会占用内存
问题：
1.线程安全——加锁解决
2.内存泄漏：无delete对象——使用共享指针

#include <iostream>
// version1:
// with problems below:
// 1. thread is not safe
// 2. memory leakclass Singleton{
private:Singleton(){std::cout<<"constructor called!"<<std::endl;}Singleton(Singleton&)=delete;Singleton& operator=(const Singleton&)=delete;static Singleton* m_instance_ptr;
public:~Singleton(){std::cout<<"destructor called!"<<std::endl;}static Singleton* get_instance(){if(m_instance_ptr==nullptr){m_instance_ptr = new Singleton;}return m_instance_ptr;}void use() const { std::cout << "in use" << std::endl; }
};Singleton* Singleton::m_instance_ptr = nullptr;int main(){Singleton* instance = Singleton::get_instance();Singleton* instance_2 = Singleton::get_instance();return 0;
}
//输出结果:constructor called!

方案2：线程安全、内存安全的懒汉式单例
问题：使用智能指针会要求用户也得使用智能指针，非必要不应该提出这种约束; 使用锁也有开销; 同时代码量也增多了，实现上我们希望越简单越好。还有更加严重的问题，在某些平台（与编译器和指令集架构有关），双检锁会失效！

#include <iostream>
#include <memory> // shared_ptr
#include <mutex>  // mutex// version 2:
// with problems below fixed:
// 1. thread is safe now
// 2. memory doesn't leakclass Singleton{
public:typedef std::shared_ptr<Singleton> Ptr;~Singleton(){std::cout<<"destructor called!"<<std::endl;}Singleton(Singleton&)=delete;Singleton& operator=(const Singleton&)=delete;static Ptr get_instance(){// "double checked lock"if(m_instance_ptr==nullptr){std::lock_guard<std::mutex> lk(m_mutex);if(m_instance_ptr == nullptr){m_instance_ptr = std::shared_ptr<Singleton>(new Singleton);}}return m_instance_ptr;}private:Singleton(){std::cout<<"constructor called!"<<std::endl;}static Ptr m_instance_ptr;static std::mutex m_mutex;
};// initialization static variables out of class
Singleton::Ptr Singleton::m_instance_ptr = nullptr;
std::mutex Singleton::m_mutex;int main(){Singleton::Ptr instance = Singleton::get_instance();Singleton::Ptr instance2 = Singleton::get_instance();return 0;
}
//输出：constructor called!
//     destructor called!

方案3：局部静态变量
解释：如果当变量在初始化的时候，并发同时进入声明语句，并发线程将会阻塞等待初始化结束。这样保证了并发线程在获取静态局部变量的时候一定是初始化过的，且只有一次初始化，所以具有线程安全性，而c++静态变量的生存期是从声明到程序结束。

#include <iostream>class Singleton
{
public:~Singleton(){std::cout<<"destructor called!"<<std::endl;}Singleton(const Singleton&)=delete;Singleton& operator=(const Singleton&)=delete;static Singleton& get_instance(){//返回引用才能获取对象static Singleton instance;return instance;}
private:Singleton(){std::cout<<"constructor called!"<<std::endl;}
};int main(int argc, char *argv[])
{Singleton& instance_1 = Singleton::get_instance();Singleton& instance_2 = Singleton::get_instance();return 0;
}
//输出：constructor called!
//     destructor called!

面试题3：数组中重复的数字

思路地址

class Solution {
public:int findRepeatNumber(vector<int>& nums) {int i = 0;while(i < nums.size()) {if(nums[i] == i) {i++;continue;}if(nums[nums[i]] == nums[i])return nums[i];swap(nums[i],nums[nums[i]]);}return -1;}
};

面试题4：二维数组中的查找

//左下角查找
class Solution {
public:bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {int i = matrix.size() - 1, j = 0;while(i >= 0 && j < matrix[0].size()){if(matrix[i][j] > target) i--;else if(matrix[i][j] < target) j++;else return true;}return false;}
};
//右上角查找
class Solution {
public:bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {if(matrix.size()==0||matrix[0].size()==0) return false;int i=0,j=matrix[0].size()-1;while(i<matrix.size()&&j>=0){if(target>matrix[i][j])++i;else if(target<matrix[i][j])--j;else return true;}return false;}
};

面试题5：替换空格

//库函数的使用
class Solution {
public:string replaceSpace(string s) {int size = s.size();for(int i=0;i<size;i++){if(s[i]==' '){s.insert(i,"%20");i+=2;s.erase(i+1,1);size+=2;}}return s;}
};
//原数组改动
class Solution {
public:string replaceSpace(string s) {
int count=0,len=s.size();for(char c:s)if(c==' ')++count;s.resize(len+2*count);for(int i=len-1,j=s.size()-1;i<j;i--,j--){if(s[i]==' '){s[j-2]='%';s[j-1]='2';s[j]='0';j-=2;}elses[j]=s[i];}return s;}
};

面试题6：从头到尾打印链表

//时间换空间：先反转链表
class Solution {
public:vector<int> reversePrint(ListNode* head) {vector<int> res;if(head==NULL) return res;ListNode* pre=NULL;ListNode* cur=head;ListNode* tmp;while(cur){tmp=cur->next;cur->next=pre;pre=cur;cur=tmp;}while(pre){res.emplace_back(pre->val);pre=pre->next;}return res;}
};
//栈的使用/**
class Solution {
public:vector<int> reversePrint(ListNode* head) {vector<int> res;stack<int> st;if(NULL==head) return res;while(head){st.emplace(head->val);head=head->next;}while(!st.empty()){res.emplace_back(st.top());st.pop();} return res;}
};

面试题7：重建二叉树

//前序和中序遍历重构二叉树
class Solution {
public:
TreeNode* helpBuildTree(
vector<int>& preorder,int  preBegin,int preEnd,
vector<int>& inorder,int inBegin,int inEnd)
{if(preBegin>=preEnd||inBegin>=inEnd) return NULL;TreeNode* root=new TreeNode(preorder[preBegin]);int index=inBegin;while(inorder[index]!=root->val){++index;}int inLeftBegin=inBegin;int inLeftEnd=index;int leftSize=inLeftEnd-inBegin;int inRightBegin=index+1;int inRightEnd=inEnd;int rightSize=inRightBegin-inRightEnd;int preLeftBegin=preBegin+1;int preLeftEnd=preLeftBegin+leftSize;int preRightBegin=preLeftEnd;int preRightEnd=preEnd;root->left=helpBuildTree(preorder,preLeftBegin,preLeftEnd,inorder,inLeftBegin,inLeftEnd);root->right=helpBuildTree(preorder,preRightBegin,preRightEnd,inorder,inRightBegin,inRightEnd);return root;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder)
{return helpBuildTree(preorder,0,preorder.size(),inorder,0,inorder.size());
}
};

面试题8：二叉树的下一个结点

/*
struct TreeLinkNode {int val;struct TreeLinkNode *left;struct TreeLinkNode *right;struct TreeLinkNode *next;TreeLinkNode(int x) :val(x), left(NULL), right(NULL), next(NULL) {}
};
*/
//next表示父节点
class Solution {
public:TreeLinkNode* GetNext(TreeLinkNode* pNode) {if (!pNode)return NULL;// 若该结点存在右子树if (pNode->right) {TreeLinkNode* rightChild = pNode->right; // 寻找右子树上的最左孩子while (rightChild->left)rightChild = rightChild->left;return rightChild;}// 若不存在右子树，寻找第一个右父亲while (pNode->next) {if (pNode->next->left == pNode) return pNode->next; pNode = pNode->next;}return NULL;}
};

面试题9：用两个栈实现队列

class CQueue {
public:CQueue() {}void appendTail(int value) {st1.push(value);}int deleteHead() {if(st1.empty()) return -1;while(!st1.empty()){st2.push(st1.top());st1.pop();}int res=st2.top();st2.pop();while(!st2.empty()){st1.push(st2.top());st2.pop();}return res;}
public:stack<int> st1;stack<int> st2;
};

面试题10-I：斐波那契数列

class Solution {
public:
int fib(int n) {if(n<=1)return n;int dp[2]={0,1};for(int i=2;i<=n;i++){long long tmp=dp[0]+dp[1];dp[0]=dp[1];dp[1]=tmp%1000000007;}//1000000007是因为题目要求要取模return dp[1];}
};

面试题10-II：青蛙跳台阶问题

class Solution {
public:int numWays(int n) {if(n==0) return 1;if(n<3) return n;int dp[2]={1,2};for(int i=3;i<=n;i++){long long tmp=dp[0]+dp[1];dp[0]=dp[1];dp[1]=tmp%1000000007;}return dp[1];}
};

面试题11：旋转数组的最小数字

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

面试题12：矩阵中的路径

class Solution {
private:int ROW, COL, LEN;bool backtracking(vector<vector<char>>& board, string& word, int row, int col, int idx) {if(idx >= LEN) {return true;}if(row < 0 || row >= ROW || col < 0 || col >= COL || board[row][col] != word[idx]) {return false;}char c = board[row][col];board[row][col] = '*';if (backtracking(board, word, row - 1, col, idx + 1) ||backtracking(board, word, row + 1, col, idx + 1) ||backtracking(board, word, row, col - 1, idx + 1) ||backtracking(board, word, row, col + 1, idx + 1)) {return true;}board[row][col] = c;return false;}
public:bool exist(vector<vector<char>>& board, string& word) {ROW = board.size();COL = board[0].size();LEN = word.length();for(int row = 0; row < ROW; ++row) {for(int col = 0; col < COL; ++col) {if(backtracking(board, word, row, col, 0)) {return true;}}}return false;}
};

面试题13：机器人的运动范围

//递归法
class Solution {
public:
int getDigitSum(int num)
{int sum=0;while(num>0){sum+=num%10;num/=10;}return sum;
}
int movingCount(int m, int n, int k)
{if(k==0) return 1;vector<vector<int>> vec(m,vector<int>(n,0));int res=1;vec[0][0]=1;for(int i=0;i<m;i++){for(int j=0;j<n;j++){if((i==0&&j==0)||getDigitSum(i)+getDigitSum(j)>k)continue;if(i>=1)vec[i][j]|=vec[i-1][j];//或运算if(j>=1)vec[i][j]|=vec[i][j-1];res+=vec[i][j];}}return res;
}
};
//递归法II
class Solution {
public:int getDigitSum(int x){int sum=0;while(x){sum+=x%10;x/=10;}return sum;}int dfs(vector<vector<int>>& dp,int m,int n,int k,int i,int j){//dp[i][j]==true表明该格子要是被访问过就继续if(i>=m||j>=n||getDigitSum(i)+getDigitSum(j)>k||dp[i][j])return 0;dp[i][j]=1;return 1+dfs(dp,m,n,k,i+1,j)+dfs(dp,m,n,k,i,j+1);}int movingCount(int m, int n, int k) {vector<vector<int>> dp(m,vector<int>(n,0));return dfs(dp,m,n,k,0,0);}
};

面试题14-I：剪绳子

class Solution {
public:
int cuttingRope(int n)
{if(n<=3)return n-1;vector<int> dp(n+1);dp[1]=1;dp[2]=2;dp[3]=3;for(int i=4;i<=n;i++){int maxValue=0;for(int j=1;j<=i/2;j++){maxValue=maxValue>dp[j]*dp[i-j]?maxValue:dp[j]*dp[i-j];}dp[i]=maxValue;}return dp[dp.size()-1];
}
};

面试题14-II：剪绳子II

//贪心算法：尽可能取3来保证最大
class Solution {
private:const long long int mod = 1e9+7;
public:int cuttingRope(int n) {if(n <= 3) return n - 1;long res = 1;while(n > 4) {res = (res * 3) % mod;n -= 3;}return (res * n) % mod;  }
};

面试题15：二进制中1的个数

class Solution {
public:
int hammingWeight(uint32_t n)
{int res=0;while(n>0){if(n%2)res++;n/=2;}return res;
}
};

面试题16：数值的整数次方

//递归
class Solution {
public:double myPow(double x, int n) {if(n==0) return 1;else if(n==-1) return 1/x;//cout<<n<<"\t"<<(n&1)<<endl;if(n&1) return myPow(x*x,n>>1)*x;//单数else return myPow(x*x,n>>1);//双数}
};

面试题17：打印从1到最大的n位数

//需要保证不溢出，需要用char或string字符串进行存储
class Solution {
public:vector<int> printNumbers(int n) {vector<int> res;int maxValue=1;maxValue=pow(10,n)for(int i=1;i<maxValue;i++)res.emplace_back(i);return res;}
};class Solution {
public:vector<int> output;vector<int> printNumbers(int n) {// 以下注释的前提：假设 n = 3if(n <= 0) return vector<int>(0);string s(n, '0'); // s最大会等于999，即s的长度为nwhile(!overflow(s)) inputNumbers(s);return output;}bool overflow(string& s){// 本函数用于模拟数字的累加过程，并判断是否越界（即 999 + 1 = 1000，就是越界情况）bool isOverFlow = false;int carry = 0; // carry表示进位for(int i=s.length()-1; i>=0; --i){int current = s[i] - '0' + carry; // current表示当前这次的操作if(i == s.length() - 1) current ++; // 如果i此时在个位，current执行 +1 操作if(current >= 10){// 假如i已经在最大的那一位了，而current++之后>=10，说明循环到头了，即999 + 1 = 1000if(i == 0) isOverFlow = true;else{// 只是普通进位，比如current从9变成10carry = 1;s[i] = current - 10 + '0'; }}else{// 如果没有进位，更新s[i]的值，然后直接跳出循环，这样就可以回去执行inputNumbers函数了，即往output里添加元素s[i] = current + '0';break;}}return isOverFlow;}void inputNumbers(string s){// 本函数用于循环往output中添加符合传统阅读习惯的元素。比如001，我们会添加1而不是001。bool isUnwantedZero = true; // 判断是否是不需要添加的0，比如001前面的两个0string temp = "";for(int i=0; i<s.length(); ++i){if(isUnwantedZero && s[i] != '0') isUnwantedZero = false;if(!isUnwantedZero) temp += s[i];}output.push_back(stoi(temp));}
};
// 作者：superkakayong
// 链接：https://leetcode-cn.com/problems/da-yin-cong-1dao-zui-da-de-nwei-shu-lcof/solution/fei-zi-jie-ti-ku-jian-17-jian-dan-da-yin-cong-1dao/

面试题18：删除链表的节点

class Solution {
public:ListNode* deleteNode(ListNode* head, int val) {
ListNode* dummyHead=new ListNode(0);dummyHead->next=head;ListNode* cur=dummyHead;while(cur->next){if(cur->next->val==val){// ListNode* tmp=cur->next;cur->next=cur->next->next;// delete tmp;break;}cur=cur->next;}cur=dummyHead->next;delete dummyHead;return cur;}
};

面试题19：正则表达式匹配

网址推荐：https://leetcode-cn.com/problems/regular-expression-matching/

//递归解法
class Solution {
public:bool isMatch(string s, string p) {return isMatch(s.c_str(), p.c_str());}bool isMatch(const char* s, const char* p) {if(*p == 0) return *s == 0;auto first_match = *s && (*s == *p || *p == '.');if(*(p+1) == '*'){return isMatch(s, p+2) || (first_match && isMatch(++s, p));}else{return first_match && isMatch(++s, ++p);}}
};
//官方题解
class Solution {
public:bool isMatch(string s, string p) {int m = s.size();int n = p.size();auto matches = [&](int i, int j) {if (i == 0) {return false;}if (p[j - 1] == '.') {return true;}return s[i - 1] == p[j - 1];};vector<vector<int>> f(m + 1, vector<int>(n + 1));f[0][0] = true;for (int i = 0; i <= m; ++i) {for (int j = 1; j <= n; ++j) {if (p[j - 1] == '*') {f[i][j] |= f[i][j - 2];if (matches(i, j - 1)) {f[i][j] |= f[i - 1][j];}}else {if (matches(i, j)) {f[i][j] |= f[i - 1][j - 1];}}}}return f[m][n];}
};

面试题20：表示数组的字符串

class Solution {
private:// 整数的格式可以用[+|-]B表示, 其中B为无符号整数bool scanInteger(const string s, int& index){if(s[index] == '+' || s[index] == '-')++index;return scanUnsignedInteger(s, index);}bool scanUnsignedInteger(const string s, int& index){int befor = index;while(index != s.size() && s[index] >= '0' && s[index] <= '9')index ++;return index > befor;}
public:// 数字的格式可以用A[.[B]][e|EC]或者.B[e|EC]表示，// 其中A和C都是整数（可以有正负号，也可以没有），而B是一个无符号整数bool isNumber(string s) {if(s.size() == 0)return false;int index = 0;//字符串开始有空格，可以返回truewhile(s[index] == ' ')  //书中代码没有该项测试++index;bool numeric = scanInteger(s, index);// 如果出现'.'，接下来是数字的小数部分if(s[index] == '.'){++index;// 下面一行代码用||的原因：// 1. 小数可以没有整数部分，例如.123等于0.123；// 2. 小数点后面可以没有数字，例如233.等于233.0；// 3. 当然小数点前面和后面可以有数字，例如233.666numeric = scanUnsignedInteger(s, index) || numeric;}// 如果出现'e'或者'E'，接下来跟着的是数字的指数部分if(s[index] == 'e' || s[index] == 'E'){++index;// 下面一行代码用&&的原因：// 1. 当e或E前面没有数字时，整个字符串不能表示数字，例如.e1、e1；// 2. 当e或E后面没有整数时，整个字符串不能表示数字，例如12e、12e+5.4numeric = numeric && scanInteger(s ,index);}//字符串结尾有空格，可以返回truewhile(s[index] == ' ')++index;return numeric && index == s.size();//最后看是否所有部分都符合，如1a3只会检测第一部分是整数然后是a就不会继续检测了，index!=size，所以返回false}
};
// 作者：san-he-n
// 链接：https://leetcode-cn.com/problems/biao-shi-shu-zhi-de-zi-fu-chuan-lcof/solution/jian-zhi-offerguan-fang-jie-da-bu-yong-c-e2wd/

面试题21：调整数组顺序使奇数位于偶数前面

//快慢指针
class Solution {
public:vector<int> exchange(vector<int>& nums) {int low = 0, fast = 0;while (fast < nums.size()) {if (nums[fast] & 1) {swap(nums[low], nums[fast]);low ++;}fast ++;}return nums;}
};
//首尾指针class Solution {
public:vector<int> exchange(vector<int>& nums) {int left = 0, right = nums.size() - 1;while (left < right){if ((nums[left] & 1) == 1){left++;continue;}if ((nums[right] & 1) == 0){right--;continue;}swap(nums[left], nums[right]);}return nums;}
};

面试题22：链表中倒数第k个节点

class Solution {
public:ListNode* getKthFromEnd(ListNode* head, int k) {ListNode* slow=head,*fast=head;while(k--&&fast){fast=fast->next;}while(fast){slow=slow->next;fast=fast->next;}return slow;}
};

面试题23：环形链表II

//快慢指针
class Solution {
public:ListNode *detectCycle(ListNode *head) {ListNode* fast = head;ListNode* slow = head;while(fast != NULL && fast->next != NULL) {slow = slow->next;fast = fast->next->next;// 快慢指针相遇，此时从head 和 相遇点，同时查找直至相遇if (slow == fast) {ListNode* index1 = fast;ListNode* index2 = head;while (index1 != index2) {index1 = index1->next;index2 = index2->next;}return index2; // 返回环的入口}}return NULL;}
};

面试题24：反转链表

class Solution {
public:ListNode* reverseList(ListNode* head) {ListNode* tmp;ListNode* cur=head;ListNode* pre=NULL;while(cur){tmp=cur->next;cur->next=pre;pre=cur;cur=tmp;}return pre;}
};

面试题25：合并两个排序的链表

class Solution {
public:ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {ListNode* dummy=new ListNode(0);ListNode* ret=dummy;while(l1&&l2){if(l1->val<l2->val){dummy->next=l1;l1=l1->next;}else {dummy->next=l2;l2=l2->next;}dummy=dummy->next;}dummy->next=(l1==NULL)?l2:l1;return ret->next;}
};

面试题26：树的子结构

class Solution {
public:bool hasSubStructure(TreeNode*A, TreeNode*B){if(!B)   return true;if(!A)return false;if(A->val!=B->val)return false;return hasSubStructure(A->left, B->left) && hasSubStructure(A->right, B->right); }bool isSubStructure(TreeNode* A, TreeNode* B) {if (!A || !B){//特殊判断return false;}   // 根节点相同的话直接进入比较，根节点不相同看B是不是A的左/右子树的子结构return hasSubStructure(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B);}
};

面试题27：二叉树的镜像

class Solution {
public:TreeNode* mirrorTree(TreeNode* root) {if (root == nullptr) {return nullptr;}TreeNode* left = mirrorTree(root->left);TreeNode* right = mirrorTree(root->right);root->left = right;root->right = left;return root;}
};

面试题28：对称的二叉树

class Solution {
public:bool traversal(TreeNode* A,TreeNode* B){if(!A&&!B) return true;else if(!A&&B) return false;else if(A&&!B) return false;else if(A->val!=B->val) return false;else return traversal(A->left,B->right)&&traversal(A->right,B->left);}bool isSymmetric(TreeNode* root) {if(!root) return true;return traversal(root->left,root->right);}
};

面试题29：顺时针打印矩阵

类似题型：54.螺旋矩阵

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {if(matrix.empty()) return vector<int>{};int up=0,left=0,right=matrix[0].size()-1,down=matrix.size()-1;vector<int> res;while(true){for(int j=left;j<=right;j++){res.emplace_back(matrix[up][j]);}if(++up>down) break;for(int i=up;i<=down;i++){res.emplace_back(matrix[i][right]);}if(--right<left) break;for(int j=right;j>=left;j--){res.emplace_back(matrix[down][j]);}if(--down<up) break;for(int i=down;i>=up;i--){res.emplace_back(matrix[i][left]);}if(++left>right) break;}return res;}
};

面试题30：包含min函数的栈

class MinStack {public:stack<int> st;stack<int> minSt;
public:/** initialize your data structure here. */MinStack() {}void push(int x) {st.push(x);if(minSt.empty())minSt.push(x);elseminSt.push(std::min(minSt.top(),x));}void pop() {st.pop();minSt.pop();}int top() {return st.top();}int min() {return minSt.top();}
};

面试题31：栈的压入、弹出序列

class Solution {
public:bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {int len=pushed.size();if(len==0) return true;stack<int> st;int indexPush=0,indexPop=0;while(true){if(st.empty()&&indexPush<len){st.push(pushed[indexPush++]);}if(popped[indexPop]==st.top()){st.pop();if(++indexPop==len)break;}else{if(indexPush<len){st.push(pushed[indexPush++]);}else{break;}}}if(indexPush==len&&indexPop==len) return true;return false;}
};

面试题32：从上到下打印二叉树

class Solution {
public:
void traversal(TreeNode* root,int depth,vector<vector<int>>& rs)
{if(!root) return;if(depth==rs.size())rs.emplace_back(vector<int>{});if(depth>rs.size())return;rs[depth].emplace_back(root->val);traversal(root->left,depth+1,rs);traversal(root->right,depth+1,rs);
}
vector<vector<int>> levelOrder(TreeNode *root)
{vector<vector<int>> rs;traversal(root,0,rs);return rs;
}
};

面试题32-III：从上到下打印二叉树II

 class Solution{
public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> ans;if(root == NULL){return ans;}queue<TreeNode*> q;q.push(root);bool isLeft = false;while(!q.empty()){int rowLen = q.size();vector<int> temp;for(int i = 0; i < rowLen; ++i){TreeNode* curNode = q.front();q.pop();if(curNode != NULL){temp.push_back(curNode->val);if(curNode->left)q.push(curNode->left);if(curNode->right)q.push(curNode->right);}}isLeft = !isLeft;if(!isLeft){ans.push_back(vector<int>(temp.rbegin(), temp.rend()));}else{ans.push_back(temp);}}return ans;}
};

面试题33：二叉搜索树的后序遍历序列

//递归法
class Solution {
public:bool traversal(const vector<int>& postorder,int left,int right)
{if(left>=right) return true;int rootVal=postorder[right];int start=left;while(start<right&&postorder[start]<rootVal) ++start;for(int i=start;i<right;++i){if(postorder[i]<=rootVal)return false;}return traversal(postorder,left,start-1)&&traversal(postorder,start,right-1);
}
bool verifyPostorder(vector<int> &postorder)
{return traversal(postorder,0,postorder.size()-1);
}
};//辅助栈
class Solution {
public:bool verifyPostorder(vector<int>& postorder) {stack<int> st;int rootVal=INT_MAX;for(auto it=postorder.rbegin();it!=postorder.rend();++it){//左子树大于root值，所以返回falseif(*it>rootVal)return false;//压入root的右节点树//*it<st.top()表示是root->right的左树节点//排空root->right->right的右树节点while(!st.empty()&&*it<st.top()){rootVal=st.top();st.pop();}//更新根节点的值为root->right->valst.push(*it);//压入}return true;}
};

面试题34：二叉树中和为某一值的路径

//递归
class Solution {
public:vector<vector<int>> res;vector<int> vec;void dfs(TreeNode* root,int target){if(!root) return;vec.push_back(root->val);target-=root->val;if(!root->left&&!root->right&&target==0){res.push_back(vec);}dfs(root->left,target);dfs(root->right,target);vec.pop_back();}vector<vector<int>> pathSum(TreeNode *root, int target){dfs(root,target);return res;}
};

面试题35：复杂链表的复制

/*
// 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];}
};

面试题36：二叉搜索树和双向链表

class Solution {
public:Node *pre, *head;Node* treeToDoublyList(Node* root) {if(root==NULL) return NULL;dfs(root);head->left=pre;pre->right=head;return head;}void dfs(Node* cur){if(cur==NULL) return;dfs(cur->left);if(pre!=NULL) pre->right=cur;else head=cur;cur->left=pre;pre=cur; dfs(cur->right);}
};

面试题37：序列化二叉树

class Codec {
public:// Encodes a tree to a single string.string serialize(TreeNode* root) {string data;queue<TreeNode*> que;if (root) que.push(root);while (!que.empty()) {auto curr = que.front();que.pop();if (curr) {data += to_string(curr->val) + ',';que.push(curr->left);que.push(curr->right);} else {data += "null,";}}if (!data.empty()) data.pop_back();return data;}// Decodes your encoded data to tree.TreeNode* deserialize(string data) {unique_ptr<TreeNode> dummy(new TreeNode(0));queue<TreeNode*> que;que.push(dummy.get());size_t beg = 0, end = 0;bool left_side = false;while (beg < data.size()) {while (end < data.size() && data[end] != ',') ++end;auto str = data.substr(beg, end - beg);TreeNode *node = nullptr;if (str != "null") node = new TreeNode(atoi(str.c_str()));auto curr = que.front();if (left_side) {curr->left = node;} else {curr->right = node;que.pop();}if (node) que.push(node);left_side = !left_side;beg = ++end;}return dummy->right;}
};

面试题38：字符串的排列

class Solution {
public:vector<string> res;string path="";void dfs(const string& s,vector<bool>& used,int index) {if(path.size()==s.size()){res.emplace_back(path);return;}for(int i=0;i<s.size();i++){if(used[i]==true){continue;}if(i>0&&used[i-1]==false&&s[i]==s[i-1]){continue;}used[i]=true;path+=s[i];dfs(s,used,index+1);path.pop_back();used[i]=false;}}vector<string> permutation(string s) {if(s.size()==0) return res;vector<bool> used(s.size(),false);sort(s.begin(),s.end());dfs(s,used,0);return res;}
};

面试题39：数组中出现次数超过一半的数字

//摩尔投票法
class Solution {
public:int majorityElement(vector<int>& nums) {int x = 0, votes = 0;for(int num : nums){if(votes == 0) x = num;votes += num == x ? 1 : -1;}return x;}
};
//链接：https://leetcode-cn.com/problems/shu-zu-zhong-chu-xian-ci-shu-chao-guo-yi-ban-de-shu-zi-lcof/solution/mian-shi-ti-39-shu-zu-zhong-chu-xian-ci-shu-chao-3/

面试题40：最小的k个数

//自定义快排
class Solution {
public:vector<int> getLeastNumbers(vector<int>& arr, int k)
{mySort(arr,0,arr.size()-1);vector<int> res;res.assign(arr.begin(),arr.begin()+k);return res;
}
private:void mySort(vector<int>& arr,int left,int right)
{if(left>=right) return;int i=left,j=right;while(i<j){while(i<j&&arr[j]>=arr[left]) --j;while(i<j&&arr[i]<=arr[left]) ++i;swap(arr[i],arr[j]);}swap(arr[i],arr[left]);//此时i=jmySort(arr,left,i-1);mySort(arr,i+1,right);
}
};
//小顶堆
class Solution {
public:vector<int> getLeastNumbers(vector<int>& arr, int k) {vector<int> res;if(k==0) return res;priority_queue<int,vector<int>,greater<int>> que;for(auto i:arr){que.push(i);}for(int i=0;i<k;i++){res.emplace_back(que.top());que.pop();}return res;}
};

面试题41：数据流中的中位数

class MedianFinder {
private:priority_queue<int,vector<int>,less<int>> small;//存储最小一半，使用大顶堆(堆顶为中位数)priority_queue<int,vector<int>,greater<int>> big;//存储最大一半，使用小顶堆(堆顶为中位数)int n=0;//n为数量
public:/** initialize your data structure here. */MedianFinder() {}void addNum(int num) {if(small.empty()){small.push(num);++n;return;}if(num<=small.top())//num比中位数小，放到small堆里{small.push(num);++n;}else{big.push(num);++n;}//small和big相差2个数字，需要重新排序if(small.size()-big.size()==2){big.push(small.top());small.pop();}if(big.size()-small.size()==2){small.push(big.top());big.pop();}}double findMedian() {if(n%2){if(small.size()>big.size()) return small.top();return big.top();}else{return ((long long)small.top() + big.top()) * 0.5;}}
};

面试题42：连续子数组的最大和

//动态规划
class Solution {
public:int maxSubArray(vector<int>& nums) {if(nums.size()<=1) return nums[0];vector<int> dp(2,0);dp[0]=nums[0];int ret=dp[0];for(int i=1;i<nums.size();i++){dp[1]=max(dp[0]+nums[i],nums[i]);ret=max(ret,dp[1]);dp[0]=dp[1];}return ret;}
};

面试题43：1~n整数中1出现的次数

class Solution {
public:int countDigitOne(int n) {int count = 0;long i = 1;//指向遍历的位数，如i=1即个位，i=10即十位，...，因为n可以有31位，所以10^31用long存储while(n/i!=0){//n/i控制遍历的次数，将所有的位数都遍历完毕long high = n/(10*i);//将当前位之前的所有高位都存在high中long cur = (n/i)%10;//将当前位记录在cur中，即我们每次都需要统计当前位上1出现的次数long low = n-(n/i)*i;if(cur == 0){count += high * i;} else if(cur == 1){count += high * i + low + 1;} else {count += high * i + i;}i = i * 10;//准备遍历下一位}return count;}
};
// 作者：lu-yang-shan-yu
// 链接：https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/c-cong-ge-wei-bian-li-dao-zui-gao-wei-yi-ci-qiu-ji/

面试题44：数字序列中某一位数字

class Solution {
public:int findNthDigit(int n) {int digit=1;//记录数字位数long long start=1;//记录n所在数字long long count=9;//记录数位个数while(n>count){n-=count;start*=10;digit++;count=9*start*digit;}int num=start+(n-1)/digit;string num_s=to_string(num);int index=(n-1)%digit;return num_s[index]-'0';}
};
//参考网址：https://leetcode-cn.com/problems/shu-zi-xu-lie-zhong-mou-yi-wei-de-shu-zi-lcof/solution/mian-shi-ti-44-shu-zi-xu-lie-zhong-mou-yi-wei-de-6/

面试题45：把数组排成最小的数

class Solution {
public:void quickSort(vector<string>& strs,int left,int right)
{if(left>=right) return;int i=left,j=right;while(i<j){while(strs[j]+strs[left]>=strs[left]+strs[j]&&i<j) --j;while(strs[i]+strs[left]<=strs[left]+strs[i]&&i<j) ++i;swap(strs[i],strs[j]);}swap(strs[left],strs[i]);quickSort(strs,left,i-1);quickSort(strs,i+1,right);
}
string minNumber(vector<int>& nums)
{vector<string> strs;string res;for(int i=0;i<nums.size();i++){strs.emplace_back(to_string(nums[i]));}quickSort(strs,0,strs.size()-1);for(string s:strs)res+=s;return res;
}
};
//lambda表达式
class Solution {
public:struct cmp{bool operator()(string s1, string s2){return s1 + s2 < s2 + s1;}};string minNumber(vector<int>& nums) {vector<string> strs;string ans;for(int i = 0; i < nums.size(); i++){strs.push_back(to_string(nums[i]));}sort(strs.begin(), strs.end(), cmp());for(int i = 0; i < strs.size(); i++){ans += strs[i];}return ans;}
};

面试题46：把数字翻译为字符串

class Solution {
public:
int translateNum(int num)
{if(num<10) return 1;//0-25编号一共26个英文字母if(num%100<10||num%100>25)//表示只能单字符return translateNum(num/10);elsereturn translateNum(num/10)+translateNum(num/100);
}
};
//参考网址：https://leetcode-cn.com/problems/ba-shu-zi-fan-yi-cheng-zi-fu-chuan-lcof/solution/cjian-ji-dai-ma-shuang-bai-by-orangeman/
//动态规划
class Solution {
public:int translateNum(int num) {// 0 - 25;//类似于跳台阶，只能一步或者两步跳string s = to_string(num);int n = s.size();vector<int>dp (n + 1, 1);//dp[0]无意义，dp[1]表示第一个数字字符for(int i = 2; i <= n; i++){int sum = (s[i - 2] - '0') * 10 + s[i - 1] - '0';if(sum >= 10 && sum <= 25) dp[i] = dp[i - 1] + dp[i - 2];else dp[i] = dp[i - 1];}return dp[n];}
};

面试题47：礼物的最大价值

//滑动窗口
class Solution {
public:int lengthOfLongestSubstring(string s) {if (s.empty())return 0;if (s.size() == 1)return 1;int res = 0, left = -1;unordered_map<char, int> hash;for (int i = 0; i < s.length(); i++){if (hash.find(s[i]) != hash.end()){left = max(hash[s[i]], left);}hash[s[i]] = i;res = max(res, i - left);}return res;}
};

面试题48：最长不含重复字符的子字符串

//滑动窗口
class Solution {
public:int lengthOfLongestSubstring(string s) {
int len=s.size();int start(0),end(0),res(0),length(0);while(end<len){char tmp=s[end];for(int index=start;index<end;index++){if(tmp==s[index]){start=index+1;length=end-start;break;}}end++;length++;res=max(res,length);}return res;}
};

面试题49：丑数

class Solution {
public:
int min(int a,int b,int c)
{a=a>b?b:a;return a>c?c:a;
}
int nthUglyNumber(int n)
{vector<int> dp(n,1);int a(0),b(0),c(0);for(int i=1;i<n;i++){dp[i]=min(dp[a]*2,dp[b]*3,dp[c]*5);if(dp[i]==(dp[a]*2)) ++a;if(dp[i]==(dp[b]*3)) ++b;if(dp[i]==(dp[c]*5)) ++c;}return dp[n-1];
}
};

面试题50：第一个只出现一次的字符

class Solution {
public:char firstUniqChar(string s) {unordered_map<char,int> umap;for(char i:s){umap[i]++;}for(int i=0;i<s.size();i++){if(umap[s[i]]==1)return s[i];}return ' ';}
};

面试题51：数组中的逆序对

//参考网址：https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/solution/jian-zhi-offer-51-shu-zu-zhong-de-ni-xu-pvn2h/
class Solution {
public:int merge_sort(int left, int right, vector<int> &nums, vector<int> &saveNums)
{if (left >= right)return 0;int mid = (left + right) >> 1;int res = merge_sort(left, mid, nums, saveNums) + merge_sort(mid + 1, right, nums, saveNums);//保存数组for (int i = left; i <= right; i++){saveNums[i] = nums[i];}int i = left;    //左集合的起点int j = mid + 1; //右集合的起点for (int k = left; k <= right; k++){if (i == mid + 1) //右集合使用完毕{nums[k] = saveNums[j++];}//左集合使用完毕或在左集合小于右集合else if (j == right + 1 || saveNums[i] <= saveNums[j]){nums[k] = saveNums[i++];}else //右集合大于左集合{nums[k] = saveNums[j++];res += mid - i + 1;}}return res;
}
int reversePairs(vector<int> &nums)
{vector<int> vec(nums.size());return merge_sort(0, nums.size() - 1, nums, vec);
}
};

面试题52：两个链表的第一个公共节点

class Solution {
public:ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {ListNode *node1 = headA;ListNode *node2 = headB;while (node1 != node2) {node1 = node1 != NULL ? node1->next : headB;node2 = node2 != NULL ? node2->next : headA;}return node1;}
};

面试题53：在排序数组中查找数字I

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

面试题53-II：0~n-1中缺失的数字

//二分法
class Solution {
public:int missingNumber(vector<int>& nums) {int left=0,right=nums.size()-1;while(left<=right){int mid=left+((right-left)>>1);if(nums[mid]==mid){left=mid+1;}else if(nums[mid]>mid){right=mid-1;}}return left;}
};
//最佳写法
class Solution {
public:int missingNumber(vector<int>& nums) {if (nums[0]==1) return 0;for (int i = 0;i<nums.size();i++){if (nums[i]!=i) return i;}return nums.size();}
};

面试题54：二叉搜索树的第K大节点

class Solution {
public:int res;void dfs(TreeNode* root,int &k){if(!root) return;dfs(root->right,k);--k;if(k==0) {res=root->val;return;}dfs(root->left,k);}int kthLargest(TreeNode* root, int k) {dfs(root,k);return res;}
};

面试题55-I：二叉树的深度

//递归
class Solution {
public:int getDepth(TreeNode* node){if(!node) return 0;int leftDepth=getDepth(node->left);int rightDepth=getDepth(node->right);int depth=1+max(leftDepth,rightDepth);return depth;}int maxDepth(TreeNode* root) {return getDepth(root);}
};

面试题55-II：平衡二叉树

class Solution {
public:int getDepth(TreeNode* root){if(!root) return 0;int left=getDepth(root->left);if(left==-1) return -1;int right=getDepth(root->right);if(right==-1) return -1;return abs(left-right)>1?-1:1+max(left,right);}bool isBalanced(TreeNode* root) {return getDepth(root)==-1?false:true;}
};

面试题56-I：数组中数字出现的次数

class Solution {
public:vector<int> singleNumbers(vector<int>& nums) {int a=0,b=0;int a_b=0,aDifb=1;//0^(异或)任何数都为原来的数//a^a=0//a_b表示最后剩下a异或b的值//aDifb表示在某一位置上a和b异或为1，及a和b在某一位置上不同，作为分界线for(int num:nums)a_b^=num;while((a_b&aDifb)==0)aDifb<<=1;//找到为1的位置for(int num:nums){if(num&aDifb)a^=num;elseb^=num;}return vector<int>{a,b};}
};

面试题56-II：数组中数字出现的次数II

class Solution
{
public:int singleNumber(vector<int> &nums){int bits[32] = {0};for (int i = 0; i < nums.size(); i++){int j = 0;while (nums[i]){bits[j++] += nums[i] % 2;nums[i] >>= 1;}}//得到二进制上各位数之和int ans = 0;//二进制装十进制并%n位数for (int i = 0; i < 32; i++){ans+=(1<<i)*(bits[i]%3);}return ans;}
};

面试题57：和为s的两个数字

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

面试题57-II：和为s的连续正数序列

class Solution {
public:vector<vector<int>> findContinuousSequence(int target) {vector<vector<int>> res;int left=1,right=1;int saveTarget=target;while(left<=(saveTarget>>1)){if(target>0){target-=right++;}else if(target<0){target+=left++;}else{vector<int> tmp;for(int i=left;i<right;i++){tmp.emplace_back(i);}res.emplace_back(tmp);target+=left++;}}return res;}
};

面试题58-I：翻转单词顺序

//头文件<sstream>
//stringstream的使用
class Solution {
public:string reverseWords(string s) {stringstream istr(s);string res,word;while(istr>>word){res=word+" "+res;}res.pop_back();return res;}
};
//双指针用法
class Solution {
public:string reverseWords(string s) {int left=s.size()-1;int right=left;string res;while(left>=0){while(s[left]==' '){--left;right=left;if(left<0) break;}if(left<0) break;while(s[left]!=' '){--left;if(left<0) break;}res+=s.substr(left+1,right-left)+" ";right=left;}return res.substr(0,res.size()-1);}
};

面试题58-II：左旋转字符串

class Solution {
public:string reverseLeftWords(string s, int n) {for(int i=0;i<n;i++){s.push_back(s[i]);}s.erase(0,n);return s;}
};

面试题59-I：滑动窗口的最大值

class Solution {
public:vector<int> maxSlidingWindow(vector<int>& nums, int k) {int n = nums.size();priority_queue<pair<int, int>> q;for (int i = 0; i < k; ++i)q.emplace(nums[i], i);vector<int> ans = { q.top().first };for (int i = k; i < n; ++i){q.emplace(nums[i], i);while (q.top().second <= i - k)q.pop();ans.push_back(q.top().first);}return ans;}
};

面试题59-II：队列的最大值

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

面试题60：n个骰子的点数

class Solution {
public:vector<double> dicesProbability(int n) {int dp[70]={0};//骰子点数和的可能方案数for(int i=1;i<=6;i++)dp[i]++;//从第二个骰子开始for(int i=2;i<=n;i++){//从点数和最大的开始for(int j=6*i;j>=i;j--){dp[j]=0;//一个骰子的点数//所有点数和范围j：[i,6*j]//除去第一个骰子后j的范围：[i-1,6*j-6]//所以下面需要判断让剩余点数(点数和-当前点数，即j-cur)<i-1for(int cur=1;cur<=6;cur++){if(j-cur<i-1) break;dp[j]+=dp[j-cur];}}}int all=pow(6,n);vector<double> ret;for(int i=n;i<=6*n;i++)ret.emplace_back(dp[i]*1.0/all);return ret;}
};

面试题61：扑克牌中的顺子

class Solution {
public:bool isStraight(vector<int>& nums) {bool card[15]={false};int minVal=14,maxVal=0;for(int num:nums){if(num==0) continue;if(card[num]) return false;card[num]=true;minVal=min(minVal,num);maxVal=max(maxVal,num);}return maxVal-minVal<5;}
};

面试题62：圆圈中最后剩下的数字

//参考网站：https://blog.csdn.net/u011500062/article/details/72855826
class Solution {
public:int lastRemaining(int n, int m) {//dp[i]：表示最后胜利的人位置，i表示当前人数int dp[n+1];//初始化dp，当人数为1时，获胜的人位置为0dp[1]=0;for(int i=2;i<=n;i++){//参考网址：https://blog.csdn.net/u011500062/article/details/72855826//解释：当人数为2时，相当于将人数为1时获胜位置0向右移动m位，//此时胜利位置为(人数为0时的胜利位置)0+m(移动位数)，//但是为了防止数组越界，需要对人数取余(%i)//综上所述，dp[i]=(dp[i-1]+m)%idp[i]=(dp[i-1]+m)%i;}return dp[n];}
};

面试题63：股票的最大利润

class Solution {
public:int maxProfit(vector<int>& prices) {int len=prices.size();if(len<=1) return 0;//dp[0][0]表示持有股票，dp[0][1]表示未持有股票vector<vector<int>> dp(2,vector<int>(2,0));dp[0][0]=-prices[0];for(int i=1;i<len;i++){dp[i%2][0]=max(dp[(i-1)%2][0],-prices[i]);dp[(i%2)][1]=max(dp[(i-1)%2][1],dp[(i-1)%2][0]+prices[i]);}return dp[(len-1)%2][1];}
};

面试题64：求1+2+…+n

class Solution {
public:int sumNums(int n) {//如果n为0返回0，然后从1开始计算n && (n += sumNums(n-1));return n;}
};

面试题65：不用加减乘除做加法

class Solution {
public:int add(int a, int b) {if (a == 0 || b == 0) {return a == 0 ? b : a;}int sum = 0, carry = 0;while (b != 0) { // 当没有进位的时候退出循环sum = a ^ b; carry = (unsigned int) (a & b) << 1; // C++ 不允许负数进行左移操作，故要加 unsigned inta = sum;b = carry;}return a;}
}
// 链接：https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof/solution/zi-jie-ti-ku-jian-65-jian-dan-bu-yong-ji-5k3q/

面试题66：构建乘积数组

class Solution {
public:vector<int> constructArr(vector<int>& a) {int n = a.size();vector<int> ret(n, 1);int left = 1;for (int i = 0; i < n; i ++) {ret[i] = left;left = left * a[i];} int right = 1;for (int i = n-1; i >= 0; i --) {ret[i] *= right;right *= a[i];}return ret;}
};
//链接：https://leetcode-cn.com/problems/gou-jian-cheng-ji-shu-zu-lcof/solution/gou-jian-cheng-ji-shu-zu-dui-cheng-bian-li-by-huwt/

面试题67：把字符串变为整数

class Solution {
public:int strToInt(string str) {int i = 0, flag = 1;int res=0;while (str[i] == ' ') i ++;if (str[i] == '-') flag = -1;if (str[i] == '-' || str[i] == '+') i ++;for (; i < str.size(); i++)  {if(str[i]<'0'||str[i]>'9') break;if(res>INT_MAX/10||(res==INT_MAX/10&&str[i]-'0'>7))return flag==1?INT_MAX:INT_MIN;res = res * 10 + (str[i] - '0');} return flag * res;}
};

面试题68-I：二叉搜索树的最近公共祖先

class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {TreeNode* ancestor = root;while (true) {if (p->val < ancestor->val && q->val < ancestor->val) {ancestor = ancestor->left;}else if (p->val > ancestor->val && q->val > ancestor->val) {ancestor = ancestor->right;}else {break;}}return ancestor;}
};

面试题69-II：二叉树的最近公共祖先

class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root==NULL||p==root||q==root){return root;}TreeNode* left=lowestCommonAncestor(root->left,p,q);TreeNode* right=lowestCommonAncestor(root->right,p,q);if(!left&&!right) return NULL;if(!left) return right;if(!right) return left;return root;// return left==NULL?right:(right==NULL?left:root);}
};

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