Leetcode/各大家 -- 10. Regular Expression Matching(DP)

10. Regular Expression Matching(DP)
hard
https://leetcode.com/problems/regular-expression-matching/

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

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思路： 见图， 1. 事先要把0presense的情况标好
                        2. 填数组，watch几个关键情况

public class Solution {
    public boolean isMatch(String s, String p) {
        boolean[][] dp = new boolean[s.length()+1][p.length()+1];
        dp[0][0] = true;
        //handle 1st row dealing a*b* - 0 presence, mark such * true
        for(int j = 1; j < dp[0].length; j++){
            if(p.charAt(j-1) == '*')
                dp[0][j] = dp[0][j-2];    
        }
        for(int i = 1; i < dp.length; i++){
            for(int j = 1; j < dp[0].length; j++){
                char sc = s.charAt(i-1);
                char pc = p.charAt(j-1);
                if(pc == sc || pc == '.'){
                    dp[i][j] = dp[i-1][j-1];
                }else if(pc == '*'){
                    if(dp[i][j-2]){
                        dp[i][j] = true;//0 presence of pre letter
                    }else if(p.charAt(j-2) == s.charAt(i-1) || p.charAt(j-2) == '.'){
                        dp[i][j] = dp[i-1][j];
                    }
                }
                //both different letter, fill false
            }
        }
        return dp[s.length()][p.length()];
    }
}