题目连接：Leetcode 044 Wildcard Matching
解题思路：动态规划，dp[i][j]表示s[0]~s[i-1]和p[0]~p[j-1]是否可以匹配。转移方程有：
s[i] == p[j] 时：dp[i][j] |= dp[i-1][j-1]
p[j] == '*'时：dp[i][j] |= dp[i-1][j] | dp[i][j-1] （分别对应*匹配0个和多个的情况）
class Solution {
public:
bool isMatch(string s, string p) {
int ns = s.size(), np = p.size();
bool** dp = new bool*[ns + 1];
for (int i = 0; i <= ns; i++) {
dp[i] = new bool[np + 1];
for (int j = 0; j <= np; j++) dp[i][j] = false;
}
dp[0][0] = true;
for (int i = 0; i <= ns; i++) {
for (int j = 1; j <= np; j++) {
if (i && (s[i-1] == p[j-1] || p[j-1] == '?' || p[j-1] == '*'))
dp[i][j] |= dp[i-1][j-1];
if (p[j-1] == '*') {
if (i) dp[i][j] |= dp[i-1][j];
dp[i][j] |= dp[i][j-1];
}
}
}
return dp[ns][np];
}
};
