LeetCode 43 Wildcard Matching

Implement wildcard pattern matching with support for

'?'

and

'*'

.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

思路：使用DP，切记在DP之前需要剪枝。
由于木有以下代码，一直TLE，呜呜呜

int count = 0;
		for (int i = 0; i < p.length(); i++) {
			if (p.charAt(i) != '*') {
				count++;
			}
		}
		if (count > s.length()) {
			return false;
		}

Accepted的代码如下

public class Solution {
	public boolean isMatch(String s, String p) {
int count = 0;
		for (int i = 0; i < p.length(); i++) {
			if (p.charAt(i) != '*') {
				count++;
			}
		}
		if (count > s.length()) {
			return false;
		}
		boolean[][] flag = new boolean[p.length() + 1][s.length() + 1];
		flag[0][0] = true;
		for (int i = 0; i < p.length(); i++) {
			if (flag[i][0] && p.charAt(i) == '*')
				flag[i + 1][0] = true;
			for (int j = 0; j < s.length(); j++) {
				if (p.charAt(i) == s.charAt(j) || p.charAt(i) == '?') {
					flag[i + 1][j + 1] = flag[i][j];
				}
				if (p.charAt(i) == '*') {
					flag[i + 1][j + 1] = flag[i + 1][j] || flag[i][j + 1];
				}
			}
		}
		return flag[p.length()][s.length()];
	}
}