LeetCode-Regular Expression Matching

Description：
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

题意：实现字符串的正则表达式的匹配，使用到的通配符有’.’和‘*’；

解法：“.”通配符可以匹配任何一个字符；“*”可以匹配零个或者多个与前一个相同的字符；因此，我们主要要考虑的就是出现“*”的情况，分为两种情况，一种另其匹配零个字符，一种另其匹配多个字符；

class Solution {
    public boolean isMatch(String s, String p) {
        if(p.length() == 0) return s.length() == 0;
        //首字符是否匹配
        boolean firstCharMatch = s.length() > 0 && (s.charAt(0) == p.charAt(0) 
                                                 || p.charAt(0) == '.');
        //存在字符'*'时
        if(p.length() >= 2 && p.charAt(1) == '*'){
            return isMatch(s, p.substring(2))//'*'匹配零个字符
                || (firstCharMatch && isMatch(s.substring(1), p));//'*'匹配多个字符
        }
        else{
            return firstCharMatch && isMatch(s.substring(1), p.substring(1));
        }
    }
}