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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 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 = “cab”
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 = “misisp*.”
Output: false

Unsolved

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(text.substring(1), pattern.substring(1));
        }
    }
}

/* 
https://leetcode-cn.com/problems/regular-expression-matching/solution/zheng-ze-biao-da-shi-pi-pei-by-leetcode/
*/

Reference
https://leetcode-cn.com/problems/regular-expression-matching