LeetCode:Edit Distance

最新推荐文章于 2025-07-23 13:59:28 发布

EddyCoding

最新推荐文章于 2025-07-23 13:59:28 发布

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分类专栏： LeetCode 文章标签： leetcode

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本文探讨了如何使用动态规划方法解决将一个字符串转换为另一个字符串所需的最少编辑步骤问题。通过分析给定的算法实现，理解了状态转移方程背后的逻辑，并提供了面试场景下的测试案例。

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Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character

这个题目的解法是动态规划。其实操作a和b本质上是一样的，从word1中删除一个字符和从word2中插入一个字符实质上没有任何区别。
dp[i][j]表示word1中前i个字符与word2中前j个字符的Edit Distance。
状态转移方程：
if(word1[i] != word2[j])
dp[i][j] = min(dp[i - 1][j], //delete one character from word1
dp[i][j - 1], //delete one character from word2
dp[i - 1][j - 1] //replace one character from both)
else
dp[i][j] = dp[i - 1][j - 1]
Solution: Dynamic Programming
Time Complexity: O(mn) Space Complexity: O(mn)

public class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1 + 1][len2 + 1];
        //Initialization
        for(int i = 1; i <= len2; i++){
            dp[0][i] = i;
        }
        for(int i = 0; i <= len1; i++){
            dp[i][0] = i;
        }
        //DP
        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; 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[len1][len2];
    }
}