【LeetCode】10. Regular Expression Matching

1.题目描述

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).

给你一个字符串s和一个字符规律p，请你来实现一个支持 ‘.’ 和 ‘*’ 的正则表达式匹配。

‘.’ 匹配任意单个字符
‘*’ 匹配零个或多个前面的那一个元素

所谓匹配，是要涵盖整个字符串s的，而不是部分字符串。

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 = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".因为 '*' 表示零个或多个，这里 'c' 为 0 个, 'a' 被重复一次。因此可以匹配字符串 "aab"。

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

2.Solutions

This Solution use 2D DP. beat 90% solutions, very simple.

Here are some conditions to figure out, then the logic can be very straightforward.

1. If p.charAt(j) == s.charAt(i) :  dp[i][j] = dp[i-1][j-1];
2. If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
3. If p.charAt(j) == '*': 
   here are two sub conditions:
   1) if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //in this case, a* only counts as empty
   2) if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
         dp[i][j] = dp[i-1][j]    //in this case, a* counts as multiple a 
      or dp[i][j] = dp[i][j-1]   // in this case, a* counts as single a
      or dp[i][j] = dp[i][j-2]   // in this case, a* counts as empty

Here is the solution

public static boolean isMatch(String s, String p) {
    if (s == null || p == null) {
        return false;
    }
    boolean[][] dp = new boolean[s.length()+1][p.length()+1];
    dp[0][0] = true;
    for (int i = 0; i < p.length(); i++) {
        if (p.charAt(i) == '*' && dp[0][i-1]) {
            dp[0][i+1] = true;
        }
    }
    for (int i = 0 ; i < s.length(); i++) {
        for (int j = 0; j < p.length(); j++) {
            if (p.charAt(j) == '.') {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == s.charAt(i)) {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == '*') {
                if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
                    dp[i+1][j+1] = dp[i+1][j-1];
                } else {
                    dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
                }
            }
        }
    }
    return dp[s.length()][p.length()];
}

（完）