44. Wildcard Matching（js）

Posted 2021-02-10 mingL

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: false
Explanation: "a" does not match the entire string "aa".

Example 2:

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

Example 3:

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

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: 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];
};