leetcode Wildcard Matching

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先把题目放上来

'?' 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") ? falseisMatch("aa","aa") ? trueisMatch("aaa","aa") ? falseisMatch("aa", "*") ? trueisMatch("aa", "a*") ? trueisMatch("ab", "?*") ? trueisMatch("aab", "c*a*b") ? false

自己用了两种方法做了这道题，无奈自己实力太差，两次都未能AC，第一次用到的是递归的思路，可是超时了，第二次选择了用动态规划的方法，同样未能AC，主要原因是空间超了，路漫漫其修远兮，吾将上下而求索，以此来勉（qi）励（pian）自己。

上代码：

bool isMatch(const char *s, const char *p) {if (*p == '\0'){return *s == '\0';}if (*s == '\0'){return true;}if (*(p + 1) == '\0'){if (*p == '*'){return true;}else if (*p == '?'){return *(s + 1) == '\0';}else{return *(s + 1) == '\0' && *p == *s;}}if (*p == '?'){return isMatch(s + 1, p + 1);}else if (*p == '*'){for (int i = 0; i < strlen(s); i++){if (isMatch(s + i, p + 1)){return true;}}return false;}else{if (*p != *s){return false;}else{return isMatch(s + 1, p + 1);}}}

动态规划：

bool isMatch(const char *s, const char *p){const int m = strlen(s);const int n = strlen(p);vector<vector<bool>> f(n + 1, vector<bool>(n + 1, false));for (int i = 0; i <= n; i++){f[i][0] = true;}for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){if (p[i - 1] == '?'){f[i][j] = f[i - 1][j - 1];}else if (p[i - 1] == s[j - 1]){f[i][j] = f[i - 1][j - 1];}else if (p[i - 1] == '*'){if (i != 1){int k;for (k = 1; k <= m; k++){if (f[i - 1][k]){break;}}for (int t = k; t <= m; t++){f[i][t] = true;}break;}else{for (int k = 1; k <= m; k++){f[1][k] = true;}}}else{return false;}}}return f[n][m];}

注意上面两个都是未能AC的。

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