WonSoon的专栏

已知：#include using namespace std;int cnt = 1;void cal(int start){    for (int i = start; i         cnt ++;        cal(i+1);    }}int main() {    cal(0);    cout     return 0

Gas StationThere are N gas stations along a circular route, where the amount of gas at stationi is gas[i].You have a car with an unlimited gas tank and it costs cost[i] of gas to trave

设矩形A（xa1，ya1），(xa2,ya2)，B（xb1,yb1），(xb2,yb2)其中：xa1xb1先看线段相交的判断：La：xa1，xa2   xa1Lb：xb1,  xb2   xb1考虑对立面（只有两种情况）：不相交的条件：xa2xb2由摩根律，得出相交条件：xa2>=xb1&&xa1矩形A B相交的条件为：分别在X和Y轴上的投影相交

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.For example,Given n = 3, your program should return all 5 unique BST's shown below. 1 3

static int prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101}; static public int mapping(String s) { int val = 1; for

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.Calling next() will return the next smallest number in the BST.Note: next() and has

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.OJ's undirected graph serialization:Nodes are labeled uniquely.We use # as a separator for each

Given a string containing only digits, restore it by returning all possible valid IP address combinations.For example:Given "25525511135",return ["255.255.11.135", "255.255.111.35"]. (Order

Sqrt(x) Implement int sqrt(int x).Compute and return the square root of x.

原码就是原来的表示方法反码是除符号位（最高位）外取反补码=反码+1以前学习二进制编码时，老师讲了一堆堆的什么原码啊反码啊补码啊xxxx转换啊，还有负数的表示方式啊 总是记不零清，终于从网上找到了一种比较好的讲解方式，保存再share一下，不过为了系统化讲解，又找来了一些编码的基础知识，如果只想看负数编码记忆法，请跳转到1.如果你不知道二进制怎么编码，请继续，否则请跳到2

二维整型矩阵Table [m][n]. 满足Table[i][j] ≤ Table[i][j + 1], Table[i][j] ≤ Table[i + 1][j]在此中进行查找元素。1.阶梯搜索从右上角或者左下角开始，如下图红线所示的查找13的过程bool stepWise(int mat[][N_MAX], int N, int targ