10. Regular Expression Matching

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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) {        if(s == null || p == null) {            return false;        }//先对数据的输入做一个简单的判断        boolean[][] dp = new boolean[s.length()+1][p.length()+1];        dp[0][0] = true;        //新建二维数组，并标记原点为true        for (int i = 0; i < p.length(); i++) {            if (p.charAt(i) == '*' && dp[0][i-1]) {                dp[0][i+1] = true;            }        }        //可能字符串一开始就是'*'，这需要单独分析。'*'可以跳两格传递true        for (int i = 0 ; i < s.length(); i++) {            for (int j = 0; j < p.length(); j++) {                if (p.charAt(j) == '.' || p.charAt(j) == s.charAt(i)) {                    dp[i+1][j+1] = dp[i][j];                }                //字符相同或者存在'.'，则dp[i+1][j+1] = dp[i][j]。                else if (p.charAt(j) == '*') {                    if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {                        dp[i+1][j+1] = dp[i+1][j-1];                        //这种情况表示'*'是要被忽略的，故跳两格传递true                    } else{                        dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);                            //这个看我写在后面的解题思路。                    }                }            }        }        return dp[s.length()][p.length()];    }}

1, If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1];
2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
3, If p.charAt(j) == '*':
              here are two sub conditions:
                              1 if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a* only counts as empty
                              2 if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
                                                 dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a
                                                  or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
                                                  or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty

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