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[LeetCode]Regular Expression Matching

时间:2014-05-15 02:34:49      阅读:388      评论:0      收藏:0      [点我收藏+]

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:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

刚开始想错了,没做对,参考网上程序写的:

class Solution {
public:
    bool isMatch(const char *s, const char *p) {
    if(*p == 0) 
	{
		return *s == 0;
	}
	//不等于‘*‘的情况
	if(*(p + 1) != ‘*‘)
	{
		//若此时*s和*p相等,则进行isMatch(s+1, p+1)的判断
		if(*s && (*s == *p || *p == ‘.‘))
		{
			return isMatch(s + 1, p + 1);
		}
		return false;
	}
	else
	{
		for(; *s && (*s == *p || *p == ‘.‘); s++)
		{
			//‘*‘ 不再匹配前面的字符.
			if(isMatch(s, p + 2))
			{
				return true;   
			}
		    //‘*‘接着匹配前面的字符.继续for循环
		}
		//‘*‘匹配结束
		return isMatch(s, p + 2);	
	}
        
    }
};


[LeetCode]Regular Expression Matching,布布扣,bubuko.com

[LeetCode]Regular Expression Matching

原文:http://blog.csdn.net/jet_yingjia/article/details/25798313

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