10. Regular Expression Matching

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 precedeng 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

class Solution {
    public boolean isMatch(String text, String pattern) {

        //递归结束条件
        if(pattern.isEmpty()) return text.isEmpty();
        //判断 text 是否为空，防止越界，如果 text 为空， 表达式直接判为 false, text.charAt(0)就不会执行了
        boolean firstmatch = (!text.isEmpty() && (pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == '.'));
        //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。
        if(pattern.length() >= 2 && pattern.charAt(1) == '*'){
            return (isMatch(text, pattern.substring(2)) || firstmatch && isMatch(text.substring(1), pattern));
        }
        else    return firstmatch && isMatch(text.substring(1), pattern.substring(1));
    }
}

https://leetcode.windliang.cc/leetCode-10-Regular-Expression-Matching.html

厉害了 我的哥