LeetCode - Edit Distance

最新推荐文章于 2021-07-03 23:03:01 发布

原创最新推荐文章于 2021-07-03 23:03:01 发布 · 246 阅读

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本文详细介绍了如何使用动态规划解决编辑距离问题，通过三维操作（插入、删除、替换），找到两个字符串之间的最小转换步骤。

https://leetcode.com/problems/edit-distance/

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

典型的DP题，用二维DP很容易就做出来了。

distance[i][j]表示word1的前i个字符到word2的前j个字符的距离。

递推公式是：

1. 如果word.charAt(i-1) == word.charAt(j-1), 那么distance[i][j] = distance[i-1][j-1]

2. 否则，就是distance[i-1][j], distance[i][j-1], distance[i][j]的最小值加1.

当i==0 或者 j==0的话，那就只能插入或删除，distance[i][j]就等于j或者i.

public class Solution {
    public int minDistance(String word1, String word2) {
        if(word1 == null || word1.length()==0) return word2==null? 0 : word2.length();
        if(word2 == null || word2.length()==0) return word1==null? 0 : word1.length();
        
        int[][] distance = new int[word1.length()+1][word2.length()+1];
        distance[0][0] = 0;
        for(int i=0; i<=word1.length(); i++){
            for(int j=0; j<=word2.length(); j++){
                if(i==0 && j==0) continue;
                if(i==0){
                    distance[i][j] = j;
                    continue;
                }
                if(j==0){
                    distance[i][j] = i;
                    continue;
                }
                if(word1.charAt(i-1)==word2.charAt(j-1)){
                    distance[i][j] = distance[i-1][j-1];
                }
                else{
                    distance[i][j] = Math.min(Math.min(distance[i-1][j], distance[i][j-1])+1, distance[i-1][j-1]+1);
                }
            }
        }
        return distance[word1.length()][word2.length()];
    }
}

不能用一维DP做哈，

distance[i][j] = Math.min(Math.min(distance[i-1][j], distance[i][j-1])+1, distance[i-1][j-1]+1);

这一句，同时用到了上一轮的j-1和这一轮的j-1，所以一维数组肯定不行，因为一维的话就只能保存一轮的distance值了。

时间复杂度和空间复杂度都是O(m*n)