ChasingTheFreeWind

class Solution {public: map<int,int> father; int Find(int x) { int r = x; while(father[r]!=r) r = father[r]; int i = x; int j; while(father[i]!=i) { j = father[i];.

并查集思路并查集，遍历数组，如果该数字+1在数组中则将它们合并。代码class Solution { unordered_map<int,int> father,cnt; int Find(int x) { int r = x; while(r!=father[r]) { r = father[r]; } int i = x; int j; .

class Solution {public: vector<int> spiralOrder(vector<vector<int>>& matrix) { vector<int> ans; int N = matrix.size(); if(N==0) return ans; int M = matrix[0].size(); int .

异或操作：任何数异或0得0，任何数异或自己得0。根据这个性质就可在时间复杂度O(N)，空间复杂度O(1)之下求到答案。class Solution {public: int singleNumber(vector<int>& nums) { int single = 0; for(int num:nums) single^=num; return single; }};...

class Solution {public: double quickPow(double x, long n) { double ans = 1.0; double mul = x; while(n) { if(n&1) ans*=mul; mul*=mul; n/=2; } r.

思路递归的方法。如果节点相等则判断子树是否相等，如果相等则返回true，反之返回false。如果节点不等，则查找其左子树和右子树，直到找到相等的节点。代码class Solution {public: bool isSame(TreeNode* s, TreeNode* t) { return (s==t)||((s!=NULL&&t!=...

思路不是很难的动态规划，但是最开始一些边界条件没考虑到，WA了好几次，然后面向测试用例debug。设dp[i]为前i天（含）旅游花费的最小金额。dp[i] = min{dp[i-1] + costs[0], dp[i-7] + costs[1], dp[i-30]+costs[2]}然后两个旅游日之间的日子旅游最小花费就等于前一个旅游日的最小花费。即dp[j] = dp[days[i]]...

官方题解代码class Solution {public: int lengthOfLongestSubstring(string s) { int N = s.length(); unordered_set<char> occ; int rk = -1; int ans = 0; for...

三次二分解决问题，注意细节就好。代码/** * // This is the MountainArray's API interface. * // You should not implement it, or speculate about its implementation * class MountainArray { * public: * int get(...

简单题可以直接用next_permutation()库函数，也可以使用dfs回溯找到所有结果。方法一 库函数class Solution {public: vector<vector<int>> permute(vector<int>& nums) { vector<vector<int> > a...

经典题目使用分治法即可class Solution {public: int merge_count(vector<int>& a) { int N = a.size(); if(N<=1) return 0; int cnt = 0; vector<int...

参考链接是个动态规划const int MOD = 1e9+7;class Solution {public: int waysToChange(int n) { vector<long long> dp; int val[4] = {25, 10, 5, 1}; dp.resize(n+1); ...