LeetCode-72.Edit Distance

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')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

时间复杂度为O(mn)

public int minDistance(String word1, String word2) {//dp mytip
        int[][] dp = new int[word1.length()+1][word2.length()+1];//word1前i个字符和word2前j个字符最少的改变次数
        for(int i=1;i<=word1.length();i++){//相当于word1和word2前同时加一个相同的字符，并初始化
            dp[i][0] = i;
        }
        for(int i=1;i<=word2.length();i++){
            dp[0][i]=i;
        }
        for(int i=0;i<word1.length();i++){
            for(int j=0;j<word2.length();j++){
                if(word1.charAt(i)==word2.charAt(j)){//如果i和j相同不需要操作
                    dp[i+1][j+1]=dp[i][j];
                }
                else{//如果不同，判断增删改，哪个操作更少
                    int min = dp[i][j+1]>dp[i+1][j]?dp[i+1][j]:dp[i][j+1];
                    min = min>dp[i][j]?dp[i][j]:min;
                    dp[i+1][j+1]=min+1;
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }

 

空间上可优化 只保存当前行和上一行

还可以适用BFS