72 Edit Distance

Leetcode

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three 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')

---

dp[i][j] 表示 word1 前 i 个字母变成 word2 需要改变的字符数，遍历 dp ，如果 word1.charAt(i-1)==word2.charAt(j-1)，word1 前 i 个字符和 word2 前 j 个字符需要变换的次数和 dp[i-1][j-1] 相同，否则可以新增、删除、更改一个字符达到 word1 前 i 个字符和 word2 前 j 个字符相同，为 dp[i-1][j-1],dp[i][j-1],dp[i-1][j] 三者最小的加一。

class Solution {
    public int minDistance(String word1, String word2) {
        int len1=word1.length(), len2=word2.length();
        int[][] dp=new int[len1+1][len2+1];
        for(int i=1;i<=len1;++i){
            dp[i][0]=i;
        }
        for(int i=1;i<=len2;++i){
            dp[0][i]=i;
        }
        for(int i=1;i<=len1;++i){
            for(int j=1;j<=len2;++j){
                if(word1.charAt(i-1)==word2.charAt(j-1)){
                    dp[i][j]=dp[i-1][j-1];
                } else {
                    dp[i][j]=1+min(dp[i-1][j-1],dp[i][j-1],dp[i-1][j]);
                }
            }
        }
        return dp[len1][len2];
    }
    
    private int min(int i1, int i2, int i3){
        if(i1<i2){
            if(i1<i3) return i1;
            return i3;
        } else {
            if(i2<i3) return i2;
            return i3;
        }
    }
}