qq_38181970的博客

class Solution {public: int rand10() { int num=(rand7()-1)*7+rand7(); while(num>10) num=(rand7()-1)*7+rand7(); return num; }};

class Solution {public: int hammingWeight(uint32_t n) { int count=0; int mask=1; for(int i=0;i<32;i++) { if(n&&mask!=0) count++; mask<<1; } ...

class Solution{public: int max_length = 0; int depth(TreeNopde *root) { if { int L = depth(root->left); int R = depth(root->right); max_length = max(max_length, L + R); ...

#include <iostream>using namespace std;#include <bits/stdc++.h>class Solution {public: vector<vector<string>> groupAnagrams(vector<string>& strs) { unordered_map<string ,vector<string> ...

class Solution {public: bool fun(int m ,int n ,int k) { int sum=0; while(m) { sum=sum+m%10; m=m/10; } while(n) { sum=sum+n%10; n=n/10; ...

class Solution {public: vector<string>v; void InOrder(TreeNode*root,vector<int>&path) { if(root) { path.push_back(root->val); //到叶子结点 if(root->left==NULL&&...

#include <iostream>#include <bits/stdc++.h>using namespace std;void BubbleSort(vector<int> &v){ for (int i = 0; i < v.size(); i++) //次数 { for (int j = 0; ...

#include &lt;stdio.h&gt;#include &lt;unistd.h&gt;#include &lt;sys/types.h&gt;#include &lt;string.h&gt;#include &lt;wait.h&gt;#include &lt;stdlib.h&gt;#include &lt;time.h&gt;int loc_1;          

      线索二叉树：(threaded binary tree)：n个结点的二叉链表中含有n+1(2n-(n-1)=n+1)个空指针域。利用二叉链表中的空指针域，存放指向结点在某种遍历次序下的前驱和后继结点的指针【注】已知二叉树的节点个数n，那么边的个数n-1。现在给出树节点的定义：class TreeNode{public:    TreeNode(int x) { v...

#include &lt;stdio.h&gt;#include &lt;unistd.h&gt;#include &lt;sys/types.h&gt;#include &lt;string.h&gt;#include &lt;wait.h&gt;#include &lt;stdlib.h&gt;#include &lt;time.h&gt;int loc_1;