LeetCode 10. Regular Expression Matching

原题链接在这里：https://leetcode.com/problems/regular-expression-matching/

题目：

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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

题解：

When it comes to two string, two array, with fixed position, asking count, max, min, true or false. It is likely to be dp.

Let dp[i][j] denotes matching result between s till index i-1 and p till index j-1.

最后返回dp[s.length()][p.length()]. 所以dp生成时要设size成new boolean[s.length()+1][p.length()+1].

状态转移时 如果当前的char能match上, dp[i][j] = dp[i-1][j-1].

否则的话如果p的当前char是'*', 要看p的前个char能否和s的当前char match上. 不能的话，这个'*'只能代表0个字符. dp[i][j] = dp[i][j-2].

如果能match上的话, 既可以代表0个字符 dp[i][j-2] 也可以代表多个字符 dp[i-1][j]. dp[i][j] = dp[i][j-2] || dp[i-1][j].

Time Complexity: O(m*n). m = s.length(), n = p.length().

Space: O(m*n).

AC Java:

public class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length();
        int n = p.length();
        boolean [][] dp = new boolean[m+1][n+1];
        dp[0][0] = true;
        
        for(int j = 1; j<=n; j++){
            if(p.charAt(j-1) == '*'){
                dp[0][j] = dp[0][j-2];
            }
        }
        
        for(int i = 1; i<=m; i++){
            for(int j = 1; j<=n; j++){
                char sChar = s.charAt(i-1);
                char pChar = p.charAt(j-1);
                //若是首字符match
                if(sChar == pChar || pChar == '.'){
                    dp[i][j] = dp[i-1][j-1];
                }else if(pChar == '*'){ //pattern 末尾是 * 
                    //pattern * 前一个char 和 pChar match, 可以是贪婪性减掉string的最后一char
                    if(sChar == p.charAt(j-2) || p.charAt(j-2) == '.'){
                        dp[i][j] = dp[i][j-2] | dp[i-1][j];
                    }else{
                        //pattern * 前一个 char 和pChar match不上，*代表0个preceding element
                        dp[i][j] = dp[i][j-2];
                    }
                }
            }
        }
        return dp[m][n];
    }
}

跟上Wildcard Matching, Edit Distance.