题解 | #计算字符串的编辑距离#

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

int minDistance(string word1, string word2) {
    int m = word1.length();
    int n = word2.length();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
    
    // 初始化边界
    for (int i = 0; i <= m; i++) {
        dp[i][0] = i;
    }
    for (int j = 0; j <= n; j++) {
        dp[0][j] = j;
    }
    
    // 动态规划
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (word1[i-1] == word2[j-1]) {
                dp[i][j] = dp[i-1][j-1];
            } else {
                dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1;
            }
        }
    }
    
    return dp[m][n];
}

int main() {
    string str1, str2;
    while (cin >> str1 >> str2) {
        cout << minDistance(str1, str2) << endl;
    }
    return 0;
}

import java.util.*;

public class Main {
    public static int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        
        // 初始化边界
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
        
        // 动态规划
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (word1.charAt(i-1) == word2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = Math.min(dp[i-1][j-1], 
                                Math.min(dp[i-1][j], dp[i][j-1])) + 1;
                }
            }
        }
        
        return dp[m][n];
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            String str1 = sc.next();
            String str2 = sc.next();
            System.out.println(minDistance(str1, str2));
        }
    }
}

def min_distance(word1: str, word2: str) -> int:
    m, n = len(word1), len(word2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    # 初始化边界
    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j
    
    # 动态规划
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if word1[i-1] == word2[j-1]:
                dp[i][j] = dp[i-1][j-1]
            else:
                dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1
    
    return dp[m][n]

# 处理多组输入
while True:
    try:
        str1 = input()
        str2 = input()
        print(min_distance(str1, str2))
    except EOFError:
        break