标签：
mplement 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
本题是匹配问题，如果遇到‘？’就匹配任意一个字符，如果是‘*’，匹配0-n个字符。利用贪心算法求解。时间：20ms。代码如下：
class Solution { public: bool isMatch(string s, string p) { if (p.empty()) return s.empty(); string::const_iterator pter = p.begin(), ster = s.begin(); string::const_iterator star_p, star_s; bool sign = false; while (ster!=s.end()){ if (pter == p.end()){ if (sign){ ster = ++star_s; pter = star_p; } else break; } else if (*pter == ‘?‘ || *pter == *ster) ++pter, ++ster; else if (*pter == ‘*‘){ while (pter != p.end() && (*pter == ‘*‘ || *pter == ‘?‘)){ if (*pter == ‘?‘){ ++pter; ++ster; if (ster == s.end()) break; } else ++pter; } if (pter==p.end()) return true; star_p = pter; star_s = ster; sign = true; } else if ((*pter != *ster || *pter!=‘?‘) && sign){ ster = ++star_s; pter = star_p; } else return false; } if (ster != s.end()) return false; while (pter!=p.end()) if (*pter++ != ‘*‘) return false; return true; } };
[LeetCode] #44 Wildcard Matching
标签：
原文地址：http://www.cnblogs.com/Scorpio989/p/4584571.html
