c0d3喵的专栏

这两道题第一眼看上去都差不多，通常思路就是用动态规划做。 对于Regular Expression Matching可以50ms通过所有test case;  而对于Wildcard Matching却需要900ms. 不知道OJ用了什么不同的test case,但是DP的时间复杂度是M*N的，没道理差别这么大。Dissuss里很多人用DP还不能通过。 原因可能如下： boo

int compareVersion(string version1, string version2) { stringstream s1(version1), s2(version2); while(1){ int i1= 0, i2 = 0; if(getline(s1, version1, '.')) //if读取成功，将字符串convert为整型 i1 = stoi(ver

Set a fast runner and a slow runner. The fast runner jumps 2 nodes at one time, and the slow runner jumps 1 nodes at one time.

Bottom up DP class Solution { public: bool wordBreak(string s, unordered_setstring> &dict) { int len = s.length(); vectorbool> dp(len + 1,false); dp[len] = true;

Recursive method: bool isMatch(const char *s, const char *p) { //p is empty if(*p == '\0') return (*s == '\0'); //p is not empty //a * is followed if(*(p + 1) == '*'){ //take 0 preceding

用STL sort重写Compare时发生了一个错误，以前没有注意到的。 我第一次写的： struct MyCmp{ bool operator()(const Interval a, const Interval b){ return (a.start } }mycmp; 在运行是得到了Runtime Error。后来把 简单测试了一下： Input： (50,60)

用了几个Character handling functions，简化了代码。挺实用的。 (ctype.h) Character handling functions This header declares a set of functions to classify and transform individual characters. Functions These fun

Something wrong with my solution as describing below. I got a wrong answer: Input: [1,2,4,4,3] Output: 10 Expected: 9 my output is 1+2+3+3+1 = 10. and I think OJ's is 1+2+3+2+1 = 9 how can

1. If want to reuse the same stringstream object, need to clear the buffer before use.

2 DP methods are used in this implementation: 1. DP method to find all palind

Implement the Manacher’s algorithm

这个笔记跟这个题目没有多大关系。

写了两种方法 1. 用STL函数 void rotate(int num[], int n, int k){ k %= n; if(k == 0) return; reverse(num, num + n); reverse(num, num + k); //[0..k-1] reverse(num + k, num +