72. Edit Distance（js）

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')

题意：给定两个单词，从第一个单词转变为第二个单词最短需要几步，只能通过增删改其中的一个字符
代码如下：

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function(word1, word2) {
     let m=word1.length,n=word2.length;
        // vector<vector<int>> dp(m+1,vector<int>(n+1,0));
    let dp=[]
    for(let i=0;i<m+1;i++){
        dp[i]=[];
        for(let j=0;j<n+1;j++){
            dp[i][j]=0;
        }
    }
        for(let i=1;i<=m;i++){
            dp[i][0]=i;
        }
        for(let i=1;i<=n;i++){
            dp[0][i]=i;
        }
        
        for(let i=1;i<=m;i++){
            for(let j=1;j<=n;j++){
                if(word1[i-1]==word2[j-1]) dp[i][j]=dp[i-1][j-1];
                else{
                    dp[i][j]=Math.min(dp[i-1][j-1]+1,Math.min(dp[i-1][j]+1,dp[i][j-1]+1));
                }
            }
        }
        return dp[m][n];
};