[Leetcode 72]编辑距离 Edit Distance

【题目】

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace ‘h‘ with ‘r‘)
rorse -> rose (remove ‘r‘)
rose -> ros (remove ‘e‘)把单词1变成单词2，可以修改/删除/插入，最少需要几步

【思路】

动态规划dp[i][j],i为原数据，j为现数据

三种情况plus、del、rep，求min+1（步骤）

【代码】

class Solution {
    public int minDistance(String word1, String word2) {
        int m=word1.length();
        int n=word2.length();
        int cost[][]=new int[m+1][n+1];
        for(int i=0;i<m;i++){
            cost[i][0]=i;}
        for(int j=0;j<n;j++){
            cost[0][j]=j;}

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(word1.charAt(i)==word2.charAt(j)){
                    cost[i+1][j+1]=cost[i][j];}
                else{
                    int plus=cost[i+1][j];
                    int del=cost[i][j+1];
                    int rep=cost[i][j];
                    cost[i+1][j+1]=Math.min(plus,Math.min(del,rep))+1;
                }
            }
        }
        return cost[m][n];
    }

}

原文地址：https://www.cnblogs.com/inku/p/10035612.html