10. Regular Expression Matching-优快云博客

本文链接：https://blog.youkuaiyun.com/csg3140100993/article/details/102705147

本文深入探讨了正则表达式匹配的实现方法，通过递归和动态规划两种方式解析了如何判断输入字符串是否完全符合给定的正则表达式模式，包括对‘.’和‘*’的支持。

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10. Regular Expression Matching

Hard

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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 (not partial).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Accepted

352,573

Submissions

1,364,685

class Solution {
public:
    
    bool isMatch(string text, string pattern) {
        if (pattern.empty()) return text.empty();
         
        bool first_match = (!text.empty() &&
                               (pattern[0] == text[0] || pattern[0] == '.'));
        //只有长度大于 2 的时候，才考虑 *
        if (pattern.length() >= 2 && pattern[1] == '*'){
            //两种情况
            //pattern 直接跳过两个字符。表示 * 前边的字符出现 0 次
            //pattern 不变，例如 text = aa ，pattern = a*，第一个 a 匹配，然后 text 的第二个 a 接着和 pattern 的第一个 a 进行匹配。表示 * 用前一个字符替代。
            return (isMatch(text, pattern.substr(2)) ||
                    (first_match && isMatch(text.substr(1), pattern)));
        } else {
            return first_match && isMatch(text.substr(1), pattern.substr(1));
        }
    }
};

class Solution {
public:
    bool isMatch(string s, string p) {
    	int m=s.length();
    	int n=p.length();
    	bool dp[m+1][n+1];
    	dp[0][0]=true;
    	for(int i=1;i<=m;i++){
    		dp[i][0]=false;//空串p只可以跟空串s匹配，其他的都为false
    	}
    	for(int j=1;j<=n;j++){//空串s可以和空串p匹配，还可以和"x*"匹配
    		if(j==1) dp[0][j]=false;
    		else{
    			if(p[j-1]=='*'){
    				dp[0][j]=dp[0][j-2];
    			}else{
    				dp[0][j]=false;
    			}
    		}
    	}
    	for(int i=1;i<=m;i++){//下面这些可以看上面的那个分析
    		for(int j=1;j<=n;j++){
    			if(p[j-1]!='*'){
    				dp[i][j]=(s[i-1]==p[j-1]||p[j-1]=='.')&&dp[i-1][j-1];
    			}else{
    				if(p[j-2]!=s[i-1]&&p[j-2]!='.'){
    					dp[i][j]=dp[i][j-2];
    				}else{
    					dp[i][j]=dp[i][j-2]||dp[i][j-1]||dp[i-1][j];
    				}
    			}
    		}
    	}
        return dp[m][n];
    }
};