README_EN.md

44. Wildcard Matching

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Description

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*' where:

• '?' Matches any single character.
• '*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

 

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'.

 

Constraints:

• 0 <= s.length, p.length <= 2000
• s contains only lowercase English letters.
• p contains only lowercase English letters, '?' or '*'.

Solutions

Python3

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True
        for j in range(1, n + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 1]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if s[i - 1] == p[j - 1] or p[j - 1] == '?':
                    dp[i][j] = dp[i - 1][j - 1]
                elif p[j - 1] == '*':
                    dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
        return dp[m][n]

Java

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

C++

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));
        dp[0][0] = true;
        for (int j = 1; j <= n; ++j) {
            if (p[j - 1] == '*') {
                dp[0][j] = dp[0][j - 1];
            }
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s[i - 1] == p[j - 1] || p[j - 1] == '?') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (p[j - 1] == '*') {
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                }
            }
        }
        return dp[m][n];
    }
};

Go

func isMatch(s string, p string) bool {
	m, n := len(s), len(p)
	dp := make([][]bool, m+1)
	for i := range dp {
		dp[i] = make([]bool, n+1)
	}
	dp[0][0] = true
	for j := 1; j <= n; j++ {
		if p[j-1] == '*' {
			dp[0][j] = dp[0][j-1]
		}
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if s[i-1] == p[j-1] || p[j-1] == '?' {
				dp[i][j] = dp[i-1][j-1]
			} else if p[j-1] == '*' {
				dp[i][j] = dp[i-1][j] || dp[i][j-1]
			}
		}
	}
	return dp[m][n]
}

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