(算法分析Week9)Wildcard Matching[Hard]
2017-11-05 13:45
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44.Wildcard Matching[Hard]
题目来源Description
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
Solution
这题和Regular Expression Matching差不多,区别在于’*’的用法,这里’*’能代替任何字符串,而不只是前一个字符的任意长度串。在这道题中,如果‘*’前面的字符没有匹配上,那么直接返回false了,根本不用继续往后搜寻。但是’*’可以匹配任意字符串,所以if (p[i - 1] == '*') dp[0][i] = dp[0][i - 1];
略微修改一下代码即可。
Complexity analysis
O(M*N)M = s.length()
N = p.length()
Code
class Solution { public: bool isMatch(string s, string p) { int M = s.size(), N = p.size(); vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false)); dp[0][0] = true; for (int i = 1; i <= N; ++i) { if (p[i - 1] == '*') dp[0][i] = dp[0][i - 1]; } for (int i = 1; i <= M; ++i) { for (int j = 1; j <= N; ++j) { if (p[j - 1] == '*') { dp[i][j] = dp[i - 1][j] || dp[i][j - 1]; } else { dp[i][j] = (s[i - 1] == p[j - 1] || p[j - 1] == '?') && dp[i - 1][j - 1]; } } } return dp[M] ; } };
Result
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