最近代码写的少了,而leetcode一直想做一个python,c/c++解题报告的专题,c/c++一直是我非常喜欢的,c语言编程练习的重要性体现在linux内核编程以及一些大公司算法上机的要求,python主要为了后序转型数据分析和机器学习,所以今天来做一个难度为hard 的简单正则表达式匹配. 做了很多leetcode题目,我们来总结一下套路: 首先一般是检查输入参数是否正确,然后是处理算法的特殊情况,之后就是实现逻辑,最后就是返回值. 当编程成为一种解决问题的习惯,我们就成为了一名纯粹的程序…
10. Regular Expression Matching Hard Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching sh…
10. Regular Expression Matching https://www.cnblogs.com/grandyang/p/4461713.html class Solution { public: bool isMatch(string s, string p) { if(p.empty()) return s.empty(); && p[] == '*') )) || (!s.empty() && (s[] == p[] || p[] == ),p)); e…
Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (n…
https://leetcode.com/problems/regular-expression-matching/ [描述] Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the ent…
Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character.'*' Matches zero or more of the preceding element. The matching should cover the entire input string (no…
Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (n…
Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be…
Implement regular expression matching with support for '.' and '*'. DP: public class Solution { public boolean isMatch2(String s, String p) { int starCnt = 0; for (int i = 0; i < p.length(); i++) { if (p.charAt(i) == '*') { starCnt++; } } boolean[] s…
问题描述: Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. http://i.cnblogs.com/EditPosts.aspx?opt=1 The matching should cover the entire input string…