leetcode 72. Edit distance edit distance (medium)

One 、 The main idea of the topic

label : Dynamic programming

https://leetcode.cn/problems/edit-distance

Here are two words for you  word1 and  word2, Please return to  word1  convert to  word2 The minimum number of operands used  .

You can do the following three operations on a word ：

Insert a character
Delete a character
Replace a character

Example  1：

Input ：word1 = "horse", word2 = "ros"
Output ：3
explain ：
horse -> rorse ( take 'h' Replace with 'r')
rorse -> rose ( Delete 'r')
rose -> ros ( Delete 'e')

Example  2：

Input ：word1 = "intention", word2 = "execution"
Output ：5
explain ：
intention -> inention ( Delete 't')
inention -> enention ( take 'i' Replace with 'e')
enention -> exention ( take 'n' Replace with 'x')
exention -> exection ( take 'n' Replace with 'c')
exection -> execution ( Insert 'u')

Tips ：

• 0 <= word1.length, word2.length <= 500
• word1 and word2 It's made up of lowercase letters

Two 、 Their thinking

Use a two-dimensional array dp[i][j], Represents the first string to position i until , And the second string to position j until , It takes at most a few steps to edit . When the first i Position and number j When the characters corresponding to bits are the same ,dp[i][j] be equal to dp[i-1][j-1]; When the corresponding characters are different , The cost of modification is dp[i-1][j-1]+1, Insert i Location / Delete j The consumption of location is dp[i][j-1]+1, Insert j Location / Delete i Location consumption is dp[i-1][j]+1.

3、 ... and 、 How to solve the problem

3.1 Java Realization

public class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0) {
                    dp[i][j] = j;
                } else if (j == 0) {
                    dp[i][j] = i;
                } else {
                    dp[i][j] = Math.min(
                        dp[i - 1][j - 1] + (word1.charAt(i - 1) == word2.charAt(j - 1) ? 0 : 1),
                        Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1)
                    );
                }
            }
        }
        return dp[m][n];
    }
}

Four 、 Summary notes

• 2022/7/4 Stick to one question every day