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", "cab") → true
思路：
用二维数组记录保存的状态dp[lens+1][lenp+1]
dp[0][0] = true表示都为空，这种情况应该返回false
1.s[i] == p[i] dp[i+1][j+1] = dp[i][j]
2.p[j] == '.' dp[i+1][j+1] = dp[i][j]
3.p[j] == ''
3.1.p[j-1] != '.' && p[j-1]!=s[i], dp[i+1][j+1] = dp[i+1][j-1] // a表示empty
3.2.
a:dp[i+1][j+1] = dp[i+1][j-1] // a表示empty
b:dp[i+1][j+1] = dp[i][j+1] // a表示多个
c:dp[i+1][j+1] = dp[i+1][j] // a表示一个，为什么是dp[i+1][j]---因为这个是最近的状态
a,b,c取或，因为是一种情况。
代码：

public boolean isMatch(String s, String p) {        
if(s == null || p == null)return false;
    
    int lenS = s.length();
    int lenP = p.length();
    
    boolean[][] dp = new boolean[lenS + 1][lenP + 1];
    dp[0][0] = true;
    
    for(int j = 0;j < lenP;j++){
        if(p.charAt(j) == '*' && dp[0][j-1]){ // p.charAt(0) != '*'
            dp[0][j+1] = true;
        }
    }
    
    for(int i=0;i<lenS;i++){
        for(int j=0;j<lenP;j++){
            if('.' == p.charAt(j)){
                dp[i+1][j+1] = dp[i][j];
            }
            
            if(s.charAt(i) == p.charAt(j)){
                dp[i+1][j+1] = dp[i][j];
            }
            
            if(p.charAt(j) == '*'){
                if(p.charAt(j-1) != '.' && p.charAt(j-1) != s.charAt(i)){
                    dp[i+1][j+1] = dp[i+1][j-1]; // a*表示匹配空
                }else {
                    dp[i+1][j+1] = (dp[i][j+1] || dp[i+1][j] || dp[i+1][j-1]); // 分别代表多个, 一个和空。
                }
            }
        }
    }
    
    return dp[lenS][lenP];
    
}