72. Edit Distance

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LeetCode

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Problem

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')

Constraints:

• $0 <= word1.length, word2.length <= 500$
• word1 and word2 consist of lowercase English letters.

Code

word1 = “horse”, word2 = “ros”

			r	o	s
		0	1	2	3
	0	0	1	2	3
h	1	1	1	2	3
o	2	2	2	1	2
r	3	3	2	2	2
s	4	4	3	3	2
e	5	5	4	4	3

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();

        int[][] dp = new int[len1 + 1][len2 + 1];
        for(int j = 1; j <= len2; j++){
            dp[0][j] = j;
        }

        for(int i = 1; i <= len1; i++){
            dp[i][0] = i;
        }

        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                char c1 = word1.charAt(i - 1);
                char c2 = word2.charAt(j - 1);

                if(c1 == c2){
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
                }
            }
        }

        return dp[len1][len2];
    }
}