leetcode这道题还挺有意思的，实现通配符，‘?‘匹配任意字符，‘*‘匹配任意长度字符串，晚上尝试了一下，题目如下：
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
直接的思路很简单，两个字符串从头开始匹配，不匹配直接返回false，匹配，则两个指针都加1，匹配子字符串，所以自然而然就想用递归来实现。
复杂一点的是*号，匹配任意长度的字符串，所以*的判断也可以循环的递归isMatch判断*后的子字符串。实现如下：
class Solution { public: bool isMatch(const char *s, const char *p) { if (‘\0‘ == *s && ‘\0‘ == *p) { return true; } if (‘*‘ == *s) { return asteriskMatch(s, p); } if (‘*‘ == *p) { return asteriskMatch(p, s); } if (*s == *p || ‘?‘ == *s || ‘?‘ == *p) { s++; p++; return isMatch(s, p); } else { return false; } } private: bool asteriskMatch(const char *asterisk, const char *p) { asterisk++; if (‘*‘ == *asterisk) { return asteriskMatch(asterisk, p); } while (*p != ‘\0‘) { if (isMatch(asterisk, p)) { return true; } p++; } if (‘\0‘ == *asterisk && ‘\0‘ == *p) { return true; } return false; } };
这样实现代码比较清爽，但是*的判断效率明显不高，提交，果然Time Limit Exceeded.没通过的用例是：
| Last executed input: | "abbabaaabbabbaababbabbbbbabbbabbbabaaaaababababbbabababaabbababaabbbbbbaaaabababbbaabbbbaabbbbababababbaabbaababaabbbababababbbbaaabbbbbabaaaabbababbbbaababaabbababbbbbababbbabaaaaaaaabbbbbaabaaababaaaabb", "**aa*****ba*a*bb**aa*ab****a*aaaaaa***a*aaaa**bbabb*b*b**aaaaaaaaa*a********ba*bbb***a*ba*bb*bb**a*b*bb" |
应该改用动态规划或者贪心法试试，不过总是觉得用递归实现比较优雅，而且看Discuss里很多人用DP也还是TLE，不行了，太困了，明天再改吧
[LeetCode]wildcard matching通配符实现TLE,布布扣,bubuko.com
[LeetCode]wildcard matching通配符实现TLE
原文：http://www.cnblogs.com/yezhangxiang/p/3915900.html