Leetcode_regular-expression-matching
2014-05-23 19:34
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http://oj.leetcode.com/problems/regular-expression-matching/
Implement regular expression matching with support for
注意这题和wildcard matching(通配符匹配)的区别。
这题比较难,也是参考了别人的代码。
用递归的方法,分当前字符的下字符是否是*两种情况,
1. 不是*的话,分两种情况,要么相等,要么*p是‘ . ’
2. 是*,先贪心的当前字符和*不匹配任何,如果递归搜索返回成功,直接返回成功,否则按字符来匹配。
具体参考代码:
Implement regular expression matching with support for
'.'and
'*'.
'.' Matches any single character. '*' Matches zero or more of the preceding element. 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", "a*") → true isMatch("aa", ".*") → true isMatch("ab", ".*") → true isMatch("aab", "c*a*b") → true
注意这题和wildcard matching(通配符匹配)的区别。
这题比较难,也是参考了别人的代码。
用递归的方法,分当前字符的下字符是否是*两种情况,
1. 不是*的话,分两种情况,要么相等,要么*p是‘ . ’
2. 是*,先贪心的当前字符和*不匹配任何,如果递归搜索返回成功,直接返回成功,否则按字符来匹配。
具体参考代码:
class Solution { public: bool isMatch(const char *s, const char *p) { if(*p=='\0') return *s=='\0'; if(*(p+1)=='*') { while((*p=='.' && *s != '\0') || (*p==*s)) { if(isMatch(s, p+2)) return true; ++s; } return isMatch(s, p+2); } else { if(*s==*p || (*p=='.' && *s!='\0')) return isMatch(s+1, p+1); return false; } } };
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