Force buckle - 10. Regular expression matching

1. subject

 

2. Ideas

First string s.length()>0, Or judge s.isEmpty();

It can be done in two steps , The first part mainly discusses the possibility of the first regular symbol , utilize &&,|| expression

situation 1： It could be a character , Look at the second symbol ;pattern.charAt(0)==text.charAt(0);

situation 2： If it is ".", You also need to see the second character ;pattern.charAt(0)=='.'

The second part is to consider the possibility of the second character .（pattern.length()>2）

situation 1： If the second character is “*”(pattern.charAt(1)=='*'), Then judge whether the character of the first regular expression is . Or characters . utilize isMatch Function to match first_match && isMatch(text.substring(1),pattern)

situation 2： If not “*”, See if it's with text The text matches first_match&&isMatch(pattern.substring(1),text.substring(1))

3. Code

public class Solution {
    public boolean isMatch(String text, String pattern) {
    	if(pattern.isEmpty()) return text.isEmpty();
        boolean first_match=(!text.isEmpty() &&(pattern.charAt(0) == text.charAt(0) || pattern.charAt(0)=='.'));
        if(pattern.length()>=2 && pattern.charAt(1)=='*'){
            return (isMatch(text,pattern.substring(2))||
            (first_match && isMatch(text.substring(1),pattern)));
        }  
        else{
            return first_match&&isMatch(pattern.substring(1),text.substring(1));
    }
}
}

Reference link ：(101 Bar message ) LeetCode：10 Regular Expression Matching （Java）_INC Follow my blog -CSDN Blog