10 Regular Expression Matching， 填dp table

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

public class Solution {
    public boolean isMatch(String s, String p) {
        // s.substring[0,i] matched p.substring[0,j]
        boolean[][] matched = new boolean[s.length() + 1][p.length() + 1];
        matched[0][0] = true;
        
        for(int j=1; j<p.length(); j += 2){
            if(p.charAt(j) == '*') {
                matched[0][j+1] = matched[0][j-1];  //matched zero before
            }
        }
        
        for(int i=1; i<=s.length(); i++){
            for(int j=1; j<=p.length(); j++){
                if(p.charAt(j-1) != '*'){
                    matched[i][j] = matched[i-1][j-1] && isCharMatch(s.charAt(i-1), p.charAt(j-1));
                } else {            
                    // 解釋清楚這兩個條件是如何推導的。
                    // i,j-2 表示match zero char,  i維持原位置，j位置的*，和j-1位置的char一同消失。
                    // i-1,j 表示match one char, s(i-1)和p(j-2)位置char是否相同.
                    matched[i][j] = matched[i][j-2] ||  
                                    matched[i-1][j] && isCharMatch(s.charAt(i-1), p.charAt(j-2));
                }
            }
        }
        
        return matched[s.length()][p.length()];
    }
    
    public boolean isCharMatch(char s, char p){
        return p == '.' || s == p;
    }
}

s = "ab"  p = ".*"
    . *
  T F T 
a F T T 
b F F T