题解 | 计算字符串的编辑距离

计算字符串的编辑距离

https://www.nowcoder.com/practice/3959837097c7413a961a135d7104c314

解题思路

编辑距离是一个经典的动态规划问题，我们可以通过以下步骤解决：

创建一个 $(m+1) \times (n+1)$ 的二维数组 $dp$ ，其中 $m$ 和 $n$ 分别是两个字符串的长度
$dp[i][j]$ 表示字符串1的前 $i$ 个字符转换到字符串2的前 $j$ 个字符所需的最小操作数
对于每个位置 $(i,j)$ ，我们有三种可能的操作：
- 替换： $dp[i][j] = dp[i-1][j-1] + (s1[i-1] \neq s2[j-1])$
- 删除： $dp[i][j] = dp[i-1][j] + 1$
- 插入： $dp[i][j] = dp[i][j-1] + 1$

代码

c++
java
python

#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));
    
    // 初始化边界条件
    for (int i = 0; i <= m; i++) {
        dp[i][0] = i;
    }
    for (int j = 0; j <= n; j++) {
        dp[0][j] = j;
    }
    
    // 填充dp数组
    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 word1, word2;
    getline(cin, word1);
    getline(cin, word2);
    cout << minDistance(word1, word2) << endl;
    return 0;
}

import java.util.Scanner;

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];
        
        // 初始化边界条件
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
        
        // 填充dp数组
        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);
        String word1 = sc.nextLine();
        String word2 = sc.nextLine();
        System.out.println(minDistance(word1, word2));
        sc.close();
    }
}

def min_distance(word1, word2):
    m, n = len(word1), len(word2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    # 初始化边界条件
    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j
    
    # 填充dp数组
    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]

word1 = input()
word2 = input()
print(min_distance(word1, word2))