10 Regular Expression Matching

Leetcode

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

'.' Matches any single character.​​​​
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).

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 (.)".  

如果当前元素匹配且下一个元素是星号，可以尝试匹配 1 个，匹配 0 个，匹配多个

class Solution {
    public boolean isMatch(String s, String p) {
        if(s==null || p==null) return false;
        return isMatch(s, p, 0, 0);
    }

    private boolean isMatch(String s, String p, int i1, int i2){
        if(i1==s.length() && i2==p.length()) return true;
        if(i2>=p.length() || i1>s.length()) return false;

        if(i2+1<p.length() && p.charAt(i2+1)=='*'){
            if(p.charAt(i2)=='.' || 
               (i1<s.length() && s.charAt(i1)==p.charAt(i2))){
                return isMatch(s, p, i1+1, i2+2) ||
                        isMatch(s, p, i1, i2+2) ||
                        isMatch(s, p, i1+1, i2);
            } else {
                return isMatch(s, p, i1, i2+2);
            }
        }

        if(i1<s.length() && s.charAt(i1)==p.charAt(i2) || p.charAt(i2)=='.'){
            return isMatch(s, p, i1+1, i2+1);
        }

        return false;
    }

}