Leetcode Proble 汇总四

第一题：

Permutations

分析：全排列问题，DFS即可。

代码：

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

第二题：

Length of Last Word

分析：水题。

代码：

class Solution
{
public:int lengthOfLastWord(const char *s) {int i, ans = 0, l1, flag;l1 = strlen(s);i = l1-1;while (s[i] == ' ') i--;for ( ; i >= 0; i--){if (s[i] != ' ') ans += 1;else break;}return ans;}
};

第三题：

Linked List Cycle

分析：设置两个指针，一个跑得快，一个跑得慢。

代码：

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution
{
public:bool hasCycle(ListNode *head) {ListNode* slow = head;ListNode* fast = head;while (fast != NULL && fast->next != NULL){slow = slow->next;fast = fast->next->next;if (slow == fast) return true;}return false;}
};

第四题：

Maximal Rectangle

分析：和以前

Largest Rectangle in Histogram

题目处理方法一样，所以复杂度是O(mn)，对每一行进行处理。

代码：

int work(vector<int> &height)
{stack<int> s;int n = height.size();int max_are = 0;int tp, are_with_top;int i = 0;while (i < n){if (s.empty() || height[s.top()] <= height[i])s.push(i++);else{tp = s.top();s.pop();are_with_top = height[tp] * (s.empty() ? i : i-s.top()-1);if (max_are < are_with_top) max_are = are_with_top;}}while (s.empty() == false){tp = s.top();s.pop();are_with_top = height[tp] * (s.empty() ? i : i-s.top()-1);if (max_are < are_with_top) max_are = are_with_top;}return max_are;
}class Solution
{
public:int maximalRectangle(vector<vector<char> > &matrix) {int size1, size2, i, j;vector<vector<int>> dp;vector<int> add;add.clear();dp.clear();size1 = matrix.size();if (size1 == 0) return 0;size2 = matrix[0].size();for (i = 0; i < size1; i++){add.clear();for (j = 0; j < size2; j++)add.push_back(0);dp.push_back(add);}for (i = 0; i < size2; i++) dp[0][i] = matrix[0][i] == '1' ? 1 : 0;for (i = 1; i < size1; i++){for (j = 0; j < size2; j++){if (matrix[i][j] == '1') dp[i][j] = dp[i-1][j]+1;else dp[i][j] = 0;}}vector<int> height;int ans = 0, ret;for (i = 0; i < size1; i++){height.clear();for (j = 0; j < size2; j++) height.push_back(dp[i][j]);ret = work(height);if (ret > ans) ans = ret;}return ans;}
};

第五题：

Linked List Cycle II

分析：如 Linked List Cycle I

代码：

/*** 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) {ListNode* slow = head;ListNode* fast = head;while (fast != NULL && fast->next != NULL){slow = slow->next;fast = fast->next->next;if (slow == fast){slow = head;while (slow != fast){slow = slow->next;fast = fast->next;}return slow;}}return NULL;}
};;

第六题：

Longest Common Prefix

分析：一开始想错了，其实比较简单吧。

代码：

class Solution
{
public:  string longestCommonPrefix(vector<string> &strs) {  string result;  int len = strs.size();  int i,j;  if (len == 0)return "";  for (j = 0; ; ++j)  {  for (i = 0;i < len; ++i)  if (j >= strs[i].length() || strs[i][j] != strs[0][j]) break;  if (i < len) break;  result += strs[0][j];  } return result;          }
};

第七题：

Binary Tree Maximum Path Sum

分析：一开始读错题了，看成必须是从叶子开始，到叶子结束。应该是从任意一个节点开始就可以==。好囧Orz。。。动态规划。

代码：

int MAX(int a, int b) { return a > b ? a : b;}
int maxnum;
void dfs(TreeNode *root)
{int a, b;if (root == NULL) return;dfs(root->left);dfs(root->right);if (root->left != NULL) a = root->left->val;else a = 0;if (root->right != NULL) b = root->right->val;else b = 0;if (root->val + a + b > maxnum) maxnum = root->val + a + b;if (root->val > maxnum) maxnum = root->val;if (a > 0 || b > 0) root->val += MAX(a, b);if (root->val > maxnum) maxnum = root->val;
}class Solution
{
public:int maxPathSum(TreeNode *root) {maxnum = -100000000;dfs(root);return maxnum;}
};

第八题：

Best Time to Buy and Sell Stock

分析：逗比了。理解错题意了。只买卖一次 =-=。 O(n)的复杂度。

代码：

int dp[101000];
class Solution
{
public:int maxProfit(vector<int> &prices) {int i, j, size, maxnum, ans = 0;memset(dp, 0, sizeof(dp));size = prices.size();if (size == 0) return 0;maxnum = size-1;dp[size-1] = prices[size-1];for (i = size-2; i >= 0; i--){if (prices[i] > prices[maxnum]) maxnum = i;dp[i] = maxnum;}maxnum = 0;for (i = 0; i < size-1; i++){if (prices[i] >= prices[i+1] && i+1 < size) continue;ans = prices[dp[i]]-prices[i];if (ans > maxnum) maxnum = ans;}return maxnum;}
};

第九题：

Best Time to Buy and Sell Stock II

分析：贪心。

代码：

class Solution
{
public:int maxProfit(vector<int> &prices) {int ans = 0, size = prices.size(), i;for (i = 1; i < size; i++)ans += prices[i]-prices[i-1] > 0 ? prices[i]-prices[i-1] : 0;return ans;}
};

第十题：

Best Time to Buy and Sell Stock III

分析：动态规划，可以参考上面第八题，开两个数组，一个用于正向DP，一个用于逆向DP，然后扫描一遍就OK了。O(n)复杂度。

代码：

int dp1[50000];
int dp2[50000];
class Solution
{
public:int maxProfit(vector<int> &prices) {int i, size;int minnum, maxnum, bestans;minnum = 100000000, maxnum = 0, bestans = 0;size = prices.size();if (size <= 1) return 0;for (i = 0; i < size; i++){if (prices[i] < minnum){minnum = prices[i];maxnum = 0;}if (prices[i] > maxnum){maxnum = prices[i];if (maxnum - minnum > bestans) bestans = maxnum-minnum;}dp1[i] = bestans;}maxnum = 0, minnum = 100000000, bestans = 0;for (i = size-1; i >= 0; i--){if (prices[i] > maxnum){maxnum = prices[i];minnum = 100000000;}if (prices[i] < minnum){minnum = prices[i];if (maxnum - minnum > bestans) bestans = maxnum - minnum;}dp2[i] = bestans;}bestans = 0;for (i = 0; i < size; i++){if (dp1[i] + dp2[i] > bestans) bestans = dp1[i]+dp2[i];}return bestans;}
};

第十一题：

Linked List Cycle

分析：两个指针，一个一次走一步，另一个一次走两步。如果有环的话，肯定能够相遇的。没有的话。。。

代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:# @param head, a ListNode# @return a booleandef hasCycle(self, head):a = headb = headif head == None:return Falsewhile True:if a.next == None or a.next.next == None:return Falsea = a.next.nextb = b.nextif a == b:return True

第十二题：

Search in Rotated Sorted Array

分析：数组是排序然后旋转的，所以利用二分可以实现O(logn)

如果A[mid] > A[left] 那么mid左边一定是有序的，同理判断右。如果无序的话，那么无非就是先升后降，或者先降后升，然后判断一下即可。

class Solution:# @param A, a list of integers# @param target, an integer to be searched# @return an integer@staticmethoddef gao(A, left, right, target):while left <= right:mid = (left + right) // 2if target == A[mid]: return midif A[mid] > A[left] and target >= A[left] and target <= A[mid]:right = mid - 1     elif A[mid] < A[right] and target >= A[mid] and target <= A[right]:left = mid + 1elif (A[mid] > target or A[left] <= target) and A[mid] < A[left]:right = mid-1else: left = mid + 1return -1def search(self, A, target):return Solution.gao(A, 0, len(A)-1, target)

第十三题：

Container With Most Water

分析：有n个高度不同的木板，他们之间的相邻距离为1，让你求用这些木板最多能够盛多少水。

贪心问题。两个指针，一个指向头pi，一个指向尾pj，如果height[pi] < height[pj], 可以分析，这已经是利用pi为边界来盛水的容量极限了。

所以就将pi指针向右移。如此即可。

class Solution:# @return an integerdef maxArea(self, height):ans = 0width, i, j = [len(height) - 1, len(height) - 1, 0]while j < i:ans = max(min(height[i], height[j]) * width , ans)width -= 1if height[i] > height[j]:j += 1else:i -= 1return ans

第十四题：

Unique Binary Search Trees

分析：直接套公式 (2n)!/(n+1)!/n!，结果要求是int,用地板除搞一下。不过貌似不用地板除也没问题。。。

代码：

class Solution:# @return an integer@staticmethoddef gao(a, b):ans = 1while a <= b:ans = ans * aa += 1return ansdef numTrees(self, n):if n < 2: return nans = Solution.gao(n+2, 2*n) // Solution.gao(1, n) return ans

第十五题：

Insertion Sort List

分析：链表的基本操作，硬伤。。。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:# @param head, a ListNode# @return a ListNodedef insertionSortList(self, head):if head == None: return Nonego = ListNode(0)ans = ListNode(0)ans.next = gogo.__init__(head.val)last = head.valit = head.nextwhile it != None:aa = ans.nextpre = ansif it.val < last:while aa != None:if aa.val > it.val:breakpre = aaaa = aa.nextpre.next = itit = it.nextpre.next.next = aaelse:last = it.valgo.next = itit = it.nextgo.next.next = Nonego = go.nextreturn ans.next

第十六题：

First Missing Positive

分析：

交换数组元素，使得数组中第i位存放数值(i+1)。最后遍历数组，寻找第一个不符合此要求的元素，返回其下标。整个过程需要遍历两次数组，复杂度为O(n)。

PS: 写过一次，但是那个是错误的=。=用了非常数大小的内存。。。所以以前那个不对。

代码：

class Solution:# @param A, a list of integers# @return an integerdef firstMissingPositive(self, A):i = 0while i < len(A):if A[i] != (i+1) and A[i] >= 1 and A[i] <= len(A) and A[A[i]-1] != A[i]:temp = A[i]A[i] = A[A[i]-1]A[temp-1] = tempelse:i += 1i = 0for i in range(len(A)):if A[i] != i+1:return i+1return len(A)+1

第十七题：

Sum Root to Leaf Numbers

分析： DFS一下即可。一开始忘记每次都要初始化ans了。。然后WA了一发。

代码：

# Definition for a  binary tree node
class TreeNode:def __init__(self, x):self.val = xself.left = Noneself.right = Noneclass Solution:# @param root, a tree node# @return an integerans = 0@staticmethoddef dfs(root, cc):cc = cc + root.valif root.left == None and root.right == None:Solution.ans = Solution.ans + ccreturnif root.left != None:Solution.dfs(root.left, 10*cc)if root.right != None:Solution.dfs(root.right, 10*cc)def sumNumbers(self, root):if root == None:return 0Solution.ans = 0Solution.dfs(root, 0)return Solution.ans

第十八题：

Set Matrix Zeroes

分析：要求把matrix中的0所在的行以及列全部元素置为0，并且原地操作。只能使用常数内存。

我们可以用matrix 的第0行以及第0列来表示matrix中的0的位置。

比如matrix[4][3]处为0，那么matrix[4][0]即为0，matrix[0][3]也为0.

这样有另一个问题，即第0行以及第0列的存在的0如何表示，我们用flag1以及flag2两个标记。来分别

表示第0行存在0，或者第0列存在0.然后扫描一下matrix即可。

class Solution:# @param matrix, a list of lists of integers# RETURN NOTHING, MODIFY matrix IN PLACE.def setZeroes(self, matrix):m, n = [len(matrix), len(matrix[0])]flag1, flag2 = [0, 0]for i in range(m):for j in range(n):                         if matrix[i][j] == 0:if i == 0 and j == 0:flag1 = 1flag2 = 1elif i == 0: flag1 = 1matrix[0][j] = 0elif j == 0: flag2 = 1matrix[i][0]else:matrix[i][0] = 0matrix[0][j] = 0i = 1while i < n:if matrix[0][i] == 0:for j in range(m):matrix[j][i] = 0i = i + 1i = 0while i < m:if matrix[i][0] == 0:for j in range(n):matrix[i][j] = 0i = i + 1if flag1 == 1:for j in range(n):matrix[0][j] = 0if flag2 == 1:for j in range(m):matrix[j][0] = 0

第十九题：

Restore IP Addresses

分析：一字符串判断是否为合法IP地址，若合法，就转化。

注意的问题：

①、如果串长度大于12，肯定不是。

②、a.b.c.d 每个数字的取值范围是 0~255

③、 0.01.0.0 这不是合法的，因为在第二个是01即若数字的长度大于1且第一位为0,，则不合法。

代码：

class Solution:# @param s, a string# @return a list of strings@staticmethoddef judge(i, j, k, s):       a, b, c, d = [0, 0, 0, 0]for m in range(i+1):a = a*10 + int(s[m])if s[0] == '0' and i > 0: a = 299for m in range(i+1, j+1): b = b*10 + int(s[m])if s[i+1] == '0' and (j-i) > 1: b = 299for m in range(j+1, k+1):c = c*10 + int(s[m])if s[j+1] == '0' and (k-j) > 1: c = 299for m in range(k+1, len(s)):d = d*10 + int(s[m])if s[k+1] == '0' and (len(s)-k-1) > 1: d = 299#if a >= 0 and a <= 255 and b >= 0 and b <= 255 and c >= 0 and c <= 255 and d >= 0 and d <= 255:ret = str(a) + '.' + str(b) + '.' + str(c) + '.' + str(d)return retreturn '-'def restoreIpAddresses(self, s):ans = []if len(s) > 12: return ansfor i in range(len(s)-3):for j in range(i+1, len(s)-2):for k in range(j+1, len(s)-1):ret = Solution.judge(i, j, k, s)if ret != '-':ans.append(ret)return ans

第二十题：

Search Insert Position

分析：二分。如果没有则返回left的值，此时left即为大于target 的第一个数的index

代码：

class Solution:# @param A, a list of integers# @param target, an integer to be inserted# @return integer@staticmethoddef binary_search(A, target):left, right = [0, len(A)-1]while left <= right:mid = (left + right) // 2if A[mid] == target:return midelif A[mid] <  target:left = mid + 1 else:right = mid - 1return leftdef searchInsert(self, A, target):return Solution.binary_search(A, target)