Leetcode72 edit Distance動态規劃

題目描述：

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:

Insert a character

Delete a character

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’)

這題的解題方法為動态規劃，java代碼如下：

class Solution {
    public int minDistance(String word1, String word2) {
        if(word1==null&&word2==null) return ;
        if(word1==null||word1.length()==){
            return word2.length();
        }else if(word2==null||word2.length()==){
            return word1.length();
        }
        //帶記事本的動态規劃
        int[][] dp=new int[word1.length()][word2.length()];
        return dynamic(word1,word2,word1.length()-,word2.length()-,dp);
    }
    public int dynamic(String word1,String word2,int top,int down,int[][] dp){
        if(top==-&&down==-) return ;
        if(top==-) return down+;
        if(down==-) return top+;
        if(dp[top][down]!=) return dp[top][down];
        if(word1.charAt(top)==word2.charAt(down)){
            dp[top][down]=dynamic(word1,word2,top-,down-,dp);
            return dp[top][down];
        }else{
            int del=dynamic(word1,word2,top,down-,dp)+;
        int rep=dynamic(word1,word2,top-,down-,dp)+;
        int add=dynamic(word1,word2,top-,down,dp)+;
        dp[top][down]=Math.min(Math.min(del,rep),add);
        return dp[top][down];
        }
    }
}