44. Wildcard Matching（js）

44. Wildcard Matching

Given an input string (s) and a pattern (p), 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).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like ? or *.

Example 1:

Input:s = "aa"p = "a"Output: falseExplanation: "a" does not match the entire string "aa".

Example 2:

Input:s = "aa"p = "*"Output: trueExplanation: ‘*‘ matches any sequence.

Example 3:

Input:s = "cb"p = "?a"Output: falseExplanation: ‘?‘ matches ‘c‘, but the second letter is ‘a‘, which does not match ‘b‘.

Example 4:

Input:s = "adceb"p = "*a*b"Output: trueExplanation: The first ‘*‘ matches the empty sequence, while the second ‘*‘ matches the substring "dce".

Example 5:

Input:s = "acdcb"p = "a*c?b"Output: false
题意：实现通配符的匹配是否合法
代码如下：

/** * @param {string} s * @param {string} p * @return {boolean} */var isMatch = function(s, p) { ???var m=s.length,n=p.length; ???var dp=[]; ???for(var i=0;i<=m;i++){ ???????dp[i]=[]; ???????for(var j=0;j<=n;j++){ ???????????dp[i][j]=false; ???????} ???} ???dp[0][0]=true; ???for(var i=1;i<=n;i++){ ???????if(p[i-1]===‘*‘) dp[0][i]=dp[0][i-1]; ???} ???????for(var i=1;i<=m;i++){ ???????for(var 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][n];};

44. Wildcard Matching（js）

原文地址：https://www.cnblogs.com/xingguozhiming/p/10424936.html