[LeetCode] 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

通配符匹配。题意是给两个字符串S和P，其中P是带有通配符的。请你返回P是否能根据通配符的规则匹配S。其中

'?' 可以匹配任何单个字符。
'*' 可以匹配任意字符串（包括空字符串）

思路是动态规划。这个题也有双指针的做法但是个人感觉动态规划的做法更好记。

首先定义动态规划的方程是dp[i][j]，定义是S的前i个字符和P的前j个字符是否相配。

初始化

• dp[0][0]:什么都没有,所以为true
• 第一行dp[0][j],换句话说,s为空,与p匹配,所以只要p开始为*才为true
• 第一列dp[i][0],当然全部为False

转移方程

• 如果(s[i] == p[j] || p[j] == "?") && dp[i-1][j-1] ,有dp[i][j] = true
• 如果p[j] == "*" && (dp[i-1][j] = true || dp[i][j-1] = true) 有dp[i][j] = true

​note:

• dp[i][j-1],表示*代表是空字符,例如ab,ab*
• dp[i-1][j],表示*代表非空任何字符,例如abcd,ab*

时间O(mn)

空间O(mn)

Java实现

class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length();
        int n = p.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 1; i <= n; i++) {
            dp[0][i] = dp[0][i - 1] && p.charAt(i - 1) == '*';
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                }
            }
        }
        return dp[m][n];
    }
}

JavaScript实现

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
    let m = s.length;
    let n = p.length;
    let dp = [...Array(m + 1)].map((item) => Array(n + 1).fill(0));
    dp[0][0] = true;
    for (let i = 1; i <= n; i++) {
        dp[0][i] = dp[0][i - 1] && p.charAt(i - 1) == '*';
    }
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (
                s.charAt(i - 1) === p.charAt(j - 1) ||
                p.charAt(j - 1) === '?'
            ) {
                dp[i][j] = dp[i - 1][j - 1];
            } else if (p.charAt(j - 1) === '*') {
                dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
            }
        }
    }
    return dp[m][n];
};

LeetCode 题目总结