题解 | 正则表达式匹配

#include <iostream>
#include <string>
#include <vector>
using namespace std;

bool isMatch(string str, string pattern) {
    int m = str.length(), n = pattern.length();
    vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
    dp[0][0] = true;
    
    // 处理pattern以*开头的情况
    for(int j = 2; j <= n; j++) {
        if(pattern[j-1] == '*') {
            dp[0][j] = dp[0][j-2];
        }
    }
    
    // 填充dp数组
    for(int i = 1; i <= m; i++) {
        for(int j = 1; j <= n; j++) {
            if(pattern[j-1] == '*') {
                dp[i][j] = dp[i][j-2];  // 匹配0次
                if(pattern[j-2] == '.' || pattern[j-2] == str[i-1]) {
                    dp[i][j] = dp[i][j] || dp[i-1][j];  // 匹配多次
                }
            }
            else if(pattern[j-1] == '.' || pattern[j-1] == str[i-1]) {
                dp[i][j] = dp[i-1][j-1];
            }
        }
    }
    
    return dp[m][n];
}

int main() {
    string str, pattern;
    cin >> str >> pattern;
    cout << (isMatch(str, pattern) ? "true" : "false") << endl;
    return 0;
}

import java.util.Scanner;

public class Main {
    public static boolean isMatch(String str, String pattern) {
        int m = str.length(), n = pattern.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        
        // 处理pattern以*开头的情况
        for(int j = 2; j <= n; j++) {
            if(pattern.charAt(j-1) == '*') {
                dp[0][j] = dp[0][j-2];
            }
        }
        
        // 填充dp数组
        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= n; j++) {
                if(pattern.charAt(j-1) == '*') {
                    dp[i][j] = dp[i][j-2];  // 匹配0次
                    if(pattern.charAt(j-2) == '.' || 
                       pattern.charAt(j-2) == str.charAt(i-1)) {
                        dp[i][j] = dp[i][j] || dp[i-1][j];  // 匹配多次
                    }
                }
                else if(pattern.charAt(j-1) == '.' || 
                        pattern.charAt(j-1) == str.charAt(i-1)) {
                    dp[i][j] = dp[i-1][j-1];
                }
            }
        }
        
        return dp[m][n];
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        String pattern = sc.nextLine();
        System.out.println(isMatch(str, pattern) ? "true" : "false");
        sc.close();
    }
}

def is_match(str, pattern):
    m, n = len(str), len(pattern)
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = True
    
    # 处理pattern以*开头的情况
    for j in range(2, n + 1):
        if pattern[j-1] == '*':
            dp[0][j] = dp[0][j-2]
    
    # 填充dp数组
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if pattern[j-1] == '*':
                dp[i][j] = dp[i][j-2]  # 匹配0次
                if pattern[j-2] == '.' or pattern[j-2] == str[i-1]:
                    dp[i][j] = dp[i][j] or dp[i-1][j]  # 匹配多次
            elif pattern[j-1] == '.' or pattern[j-1] == str[i-1]:
                dp[i][j] = dp[i-1][j-1]
    
    return dp[m][n]

str = input()
pattern = input()
print("true" if is_match(str, pattern) else "false")