Leetcode 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

這個題的題意與劍指Offer中一個題目類似，均為模擬一個正規表示式的匹配過程，即為字串的匹配問題，這類問題使用dp可以解決。

class Solution {
    public boolean isMatch(String s, String p) {
        //陣列標識的為元素的實際位置
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[0][0] = true;
        //判斷“a”和“a*”match的情況
        for(int i = 1 ; i <= p.length() ;i++){
            if(p.charAt(i - 1) == '*' && dp[0][i - 2]){
                dp[0][i] = true;
            }
        }
        for(int i = 1 ; i <= s.length();i++){
            for(int j = 1 ; j <= p.length();j++){
                //單個字元匹配的情況
                if(s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.'){
                    dp[i][j] = dp[i - 1][j - 1];
                }else if(p.charAt(j - 1) == '*'){//字元未匹配，使用*來進行填補
                    //當字元沒有匹配而且也沒有“.”的情況
                    if(p.charAt(j - 2) != s.charAt(i - 1) && p.charAt(j - 2) != '.'){
                        dp[i][j] = dp[i][j - 2];
                    }else{
                        //可以進行匹配的三個條件
                        dp[i][j] = (dp[i][j - 2] || dp[i][j - 1] || dp[i - 1][j]);
                    }
                }
            }
        }
        //返回最後dp的值，如果一路匹配下來則為true，但是如果在其中任何一個點斷了就為false
        return dp[s.length()][p.length()];
    }
}

接下來為奇招：

class Solution {
    public boolean isMatch(String s, String p) {
        return s.matches(p);   
    }
}

直接套用Regex的庫函式進行判斷。

推薦：正規表示式

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