leetcode 44. Wildcard Matching

难度：hard

题目：

与leetcode10. Regular Expression Matching此题类似，同样采用动态规划，这两题解法很精彩，比较经典

AC解：

class Solution {public:    bool isMatch(string s, string p) {        int slen = s.length();        int plen = p.length();        if (slen == 0 && plen == 0)            return true;        vector<vector<bool>> dp(slen + 1, vector<bool>(plen + 1, false));        //初始化数组        dp[0][0] = true;        for (int i = 0; i < plen && p[i] == '*'; i++)            dp[0][i + 1] = true;        for (int i = 1; i < slen + 1; i++)            for (int j = 1; j < plen + 1; j++) {                if (s[i - 1] == p[j - 1] || p[j - 1] == '?')                    dp[i][j] = dp[i - 1][j - 1];                if (p[j - 1] == '*') {                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];                }            }        return dp[slen][plen];    }};