README_EN.md

10. Regular Expression Matching

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Description

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

• '.' Matches any single character.​​​​
• '*' Matches zero or more of the preceding element.

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 = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input: s = "aab", p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".

Example 5:

Input: s = "mississippi", p = "mis*is*p*."
Output: false

 

Constraints:

• 0 <= s.length <= 20
• 0 <= p.length <= 30
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

Solutions

Python3

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        if n == 0:
            return m == 0
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True
        for j in range(2, n + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 2]
        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] == '*':
                    if p[j - 2] == '.' or p[j - 2] == s[i - 1]:
                        dp[i][j] = dp[i][j - 2] or dp[i - 1][j]
                    else:
                        dp[i][j] = dp[i][j - 2]
        return dp[-1][-1]

Java

class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        if (n == 0) {
            return m == 0;
        }
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int j = 1; j < n + 1; ++j) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 2];
            }
        }
        for (int i = 1; i < m + 1; ++i) {
            for (int j = 1; j < n + 1; ++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) == '*') {
                    if (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') {
                        dp[i][j] = dp[i][j - 2] || dp[i - 1][j];
                    } else {
                        dp[i][j] = dp[i][j - 2];
                    }
                }
            }
        }
        return dp[m][n];
    }
}

C++

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        if (n == 0) return m == 0;
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
        dp[0][0] = true;
        for (int j = 1; j < n + 1; ++j) {
            if (p[j - 1] == '*') {
                dp[0][j] = dp[0][j - 2];
            }
        }
        for (int i = 1; i < m + 1; ++i) {
            for (int j = 1; j < n + 1; ++j) {
                if (s[i - 1] == p[j - 1] || p[j - 1] == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (p[j - 1] == '*') {
                    if (s[i - 1] == p[j - 2] || p[j - 2] == '.') {
                        dp[i][j] = dp[i][j - 2] || dp[i - 1][j];
                    } else {
                        dp[i][j] = dp[i][j - 2];
                    }
                }
            }
        }
        return dp[m][n];
    }
};

Go

func isMatch(s string, p string) bool {
    m, n := len(s), len(p)
    if n == 0 {
        return m == 0
    }
    dp := make([][]bool, m + 1)
    for i := 0; i < m + 1; i++ {
        dp[i] = make([]bool, n + 1)
    }
    dp[0][0] = true
    for j := 1; j < n + 1; j++ {
        if p[j - 1] == '*' {
            dp[0][j] = dp[0][j - 2]
        }
    }
    for i := 1; i < m + 1; i++ {
        for j := 1; j < n + 1; j++ {
            if s[i - 1] == p[j - 1] || p[j - 1] == '.' {
                dp[i][j] = dp[i - 1][j - 1]
            } else if p[j - 1] == '*' {
                if s[i - 1] == p[j - 2] || p[j - 2] == '.' {
                    dp[i][j] = dp[i][j - 2] || dp[i - 1][j]
                } else {
                    dp[i][j] = dp[i][j - 2]
                }
            }
        }
    }
    return dp[m][n]
}

JavaScript

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
    let memo = {};
    function recursive(i, j) {
        if (memo[[i, j]] !== undefined) return memo[[i, j]];
        if (j === p.length) return i === s.length;
        let tmp = i < s.length && (s[i] === p[j] || p[j] === ".");
        let ans = false;
        if (p[j + 1] === "*") {
            ans = recursive(i, j + 2) || (tmp && recursive(i + 1, j));
        } else {
            ans = tmp && recursive(i + 1, j + 1);
        }
        memo[[i, j]] = ans;
        return ans;
    }
    return recursive(0, 0);
};

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