LeetCode : Regular Expression Matching-优快云博客

本文链接：https://blog.youkuaiyun.com/zero0914/article/details/32380985

本文详细阐述了如何实现正则表达式的匹配功能，特别关注支持'.'和'*'符号的用法。通过动态规划方法解决匹配问题，并解释了不同情况下的时间复杂度。

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From : https://oj.leetcode.com/problems/regular-expression-matching/

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

Let us rule the length of string s is l_s,and the length of string p is l_p;

It is easy to come up with an idea using Dynamic Programming, the time complexity of which is O(l_p*l_s). Use this method to solve this problem, and Time Limit Exceeded will be given by Online Judge.

Notice that "*" can match zero or more of the preceding element, the time complexity of enumeration is also O(l_s*l_p), but for the elements in string p which are not '*', the time complexity of transfer will be O(1), which is the reason why enumeration use less time.

class Solution {
public:
    bool isMatch(const char *s, const char *p){
        if(!p[0]) return !s[0];
        if(p[1]=='*'){  
            while(s[0]&&(s[0]==p[0]||p[0]=='.')){  
                if(isMatch(s,p+2)) return true;
                s++;                  
            }
            return isMatch(s,p+2);  
        }
        else if(s[0]&&(s[0]==p[0]||p[0]=='.'))return isMatch(s+1,p+1);
        return false; 
    }
};