leetcode 010 —— Regular Expression Matching(hard)
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题目:
'.' 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true
题意:
相关题目:
Wildcard Matching 44
思路1:递归实现 (效率很低,花了整整480ms)
如果P[j+1]!='*',S[i] == P[j] 匹配下一位(i+1, j+1),S[i]!=P[j] 匹配失败;
如果P[j+1]=='*',S[i]==P[j] 匹配(i, j+2),S[i]!=P[j] 匹配下一位(i,j+2)。
匹配成功的条件为S[i]=='\0' && P[j]=='\0'。
class Solution {public:bool isMatch(string s, string p) {if (p.size()==0) return s.size()==0;if (p.size() == 1)return (s.size() == 1) && (s[0] == p[0] || p[0] == '.');if (p[1] != '*'){if (s.size()&&(s[0] == p[0] || p[0] == '.'))return isMatch(s.substr(1), p.substr(1));elsereturn false;}while (s.size()&&(s[0] == p[0] || p[0] == '.')){if (isMatch(s, p.substr(2)))return true;s = s.substr(1);}return isMatch(s, p.substr(2));}};
思路二:动态规划:
以现在的水平还不足以领会这个问题,以后回来解决
http://blog.csdn.net/fzzying3/article/details/42057935
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