X1579983102的博客

class Solution {public: int move(vector<vector<int>>& o) { int length = o.size(), width = o[0].size(); if(length==0|| width==0||o[0][0] == 1) return 0; vector<vector<int>> temp(length + 1, vector&l..

动态规划:class Solution {public: int maxProduct(vector<int>& nums) { int len = nums.size(), ans = nums[0]; int prevMin = nums[0], prevMax = nums[0]; int temp1 = 0, temp2 = 0; for (int i = 1; i < len; i++) { temp1 = prevMin.

前缀和解法:class Solution {public: int subarraySum(vector<int>& nums, int k) { int count = 0; for (int i = 0; i < nums.size(); i++) { //实现从任意的num[i]加到[j] int sum = 0; for (int j = i; j>=0; j--) { .

异或运算符解法：class Solution {public: int singleNumber(vector<int>& nums) { int res=0; for(int i=0;i<nums.size();i++) res^=nums[i]; //^:异或运算符号 return res; }};原理：相同的数字两两相消，最后只有单身贵族活着另解：class Solution {.

/** * 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>&gt.

递归二分法算x的n次方class Solution {public: double myPow(double x, long n) { if(x==-1){ if(n%2==0) return 1; else return -1; }else{ if(n==0||x==1) return 1; if(n==1) return x; if(n&l

重点是getMin()，整块代码为此服务，其他的都能直接调用`class MinStack {public: stack<int>S; stack<int>Min; //记录每次入栈后变化的最小值 void push(int x) { S.push(x); if(Min.empty()||x<=Min.top())Min.push(x); } void pop() { if.