044 Wildcard Matching 044 Wildcard Matching 这道题直观想法是用dp 复杂度 O(mn) 代码如下, 但是在用python的时候会超时 classSolution:#@param {string} s#@param {string} p#@return {boolean}defisMatch(self, s, p): m, n=len(p), len(s) dp= [Falseforiinrange(0, n+1)] dp[0]=True...
原题地址:https://oj.leetcode.com/problems/wildcard-matching/ 题意: 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 ...
class Solution { public: bool isMatch(string s, string p) { int m = s.length(), n = p.length(); int i = 0, j = 0, asterisk = -1, match; while (i < m) { if (j < n && p[j] == '*') { match = i; asterisk = j++; } else if (j < n && (s[i] == p[j...
AI代码解释 classSolution{public:boolisMatch(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...
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 ...
Python version Windows machine: ('python=3.8.10 (tags/v3.8.10:3d8993a, May 3 2021, 11:48:03) [MSC v.1928 64 bit (AMD64)]') 'os=Windows-10-10.0.19042-SP0' 'numpy=1.23.3' 'asammdf=7.1.1' Linux machine: 'python=3.8.13 (default, Mar 16 2022,...
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