README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3407.Substring%20Matching%20Pattern/README_EN.md	String String Matching

3407. Substring Matching Pattern

中文文档

Description

You are given a string s and a pattern string p, where p contains exactly one '*' character.

The '*' in p can be replaced with any sequence of zero or more characters.

Return true if p can be made a substring of s, and false otherwise.

 

Example 1:

Input: s = "leetcode", p = "ee*e"

Output: true

Explanation:

By replacing the '*' with "tcod", the substring "eetcode" matches the pattern.

Example 2:

Input: s = "car", p = "c*v"

Output: false

Explanation:

There is no substring matching the pattern.

Example 3:

Input: s = "luck", p = "u*"

Output: true

Explanation:

The substrings "u", "uc", and "uck" match the pattern.

 

Constraints:

• 1 <= s.length <= 50
• 1 <= p.length <= 50
• s contains only lowercase English letters.
• p contains only lowercase English letters and exactly one '*'

Solutions

Solution 1

Python3

class Solution:
    def hasMatch(self, s: str, p: str) -> bool:
        i = 0
        for t in p.split("*"):
            j = s.find(t, i)
            if j == -1:
                return False
            i = j + len(t)
        return True

Java

class Solution {
    public boolean hasMatch(String s, String p) {
        int i = 0;
        for (String t : p.split("\\*")) {
            int j = s.indexOf(t, i);
            if (j == -1) {
                return false;
            }
            i = j + t.length();
        }
        return true;
    }
}

C++

class Solution {
public:
    bool hasMatch(string s, string p) {
        int i = 0;
        int pos = 0;
        int start = 0, end;
        while ((end = p.find("*", start)) != string::npos) {
            string t = p.substr(start, end - start);
            pos = s.find(t, i);
            if (pos == string::npos) {
                return false;
            }
            i = pos + t.length();
            start = end + 1;
        }
        string t = p.substr(start);
        pos = s.find(t, i);
        if (pos == string::npos) {
            return false;
        }
        return true;
    }
};

Go

func hasMatch(s string, p string) bool {
	i := 0
	for _, t := range strings.Split(p, "*") {
		j := strings.Index(s[i:], t)
		if j == -1 {
			return false
		}
		i += j + len(t)
	}
	return true
}

TypeScript

function hasMatch(s: string, p: string): boolean {
    let i = 0;
    for (const t of p.split('*')) {
        const j = s.indexOf(t, i);
        if (j === -1) {
            return false;
        }
        i = j + t.length;
    }
    return true;
}