1 题目: 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 sho…

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 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…

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…

一.题目说明 这个题目是10. Regular Expression Matching,乍一看不是很难. 但我实现提交后,总是报错.不得已查看了答案. 二.我的做法 我的实现,最大的问题在于对.*的处理有问题,始终无法成功. #include<iostream> using namespace std; class Solution{ public: bool isMatch(string s,string p){ bool result = true; if(s.length()<=0…

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 '*'. '.' 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…

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 '*'. 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…

题目描述: Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. 解题思路: 这道题如果只考虑“.”的话其实很好完成,所以解题的关键在于处理“*”的情况.以为“*”与前一个字母有关,所以应该整体考虑ch*……的情况.ch*可以匹配0-n个s的字符串…