LeetCode 72 questions: edit distance (dynamic programming)

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

Status: dp[i][j]said it would word1[0, i]be converted into word2[0, j]the number of programs.

State transition equation: Category talk (from the middle to start thinking about the situation).

(1) If word1.charAt(i) == word2.charAt(j)true, then do dp[i][j] = dp[i - 1][j - 1]nothing, ;
(2) otherwise:

① modifications word1[i]become word2[j]; after
② the word1[0, i]future of the last character to delete;
③ will word1[0, i]be added after the end of a character;

Converted from the three cases.

dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1

• Initialization: Here the array dpto open a line, to open one, in order to satisfy the boundary conditions.

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

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

• Output:dp[len1][len2]

• State Reduction: just look at the current line on the line, so you can use the "scroll array" or compressed directly into a row (difficult, therefore, not compressed).

Java code:

public class Solution4 {

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

        int[][] dp = new int[len1 + 1][len2 + 1];
        
        // 初始化
        for (int i = 0; i <= len1; i++) {
            dp[i][0] = i;
        }

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

        for (int i = 0; i < len1; i++) {
            for (int j = 0; j < len2; j++) {
                if (word1.charAt(i) == word2.charAt(j)) {
                    dp[i + 1][j + 1] = dp[i][j];
                    continue;
                }
                dp[i + 1][j + 1] = Math.min(dp[i][j + 1], Math.min(dp[i + 1][j], dp[i][j])) + 1;
            }
        }
        return dp[len1][len2];
    }
}

Really do it "state Compression", I found not so easy, because of initialization.

Java code:

import java.util.Arrays;

public class Solution5 {

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

        int[][] dp = new int[len1 + 1][len2 + 1];

        // 状态转移方程：dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1
        // 初始化

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

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


        // 打印输出
        for (int i = 0; i < dp.length; i++) {
            System.out.println(Arrays.toString(dp[i]));
        }
        System.out.println();

        for (int i = 0; i < len1; i++) {
            for (int j = 0; j < len2; j++) {
                if (word1.charAt(i) == word2.charAt(j)) {
                    dp[i + 1][j + 1] = dp[i][j];
                    continue;
                }
                dp[i + 1][j + 1] = Math.min(dp[i][j + 1], Math.min(dp[i + 1][j], dp[i][j])) + 1;
            }
        }

        // 打印输出
        for (int i = 0; i < dp.length; i++) {
            System.out.println(Arrays.toString(dp[i]));
        }

        return dp[len1][len2];
    }


    public static void main(String[] args) {
        Solution5 solution5 = new Solution5();
        String word1 = "horse";
        String word2 = "ros";
        int res = solution5.minDistance(word1, word2);
        System.out.println(res);
    }
}

Output:

[0, 1, 2, 3]
[1, 0, 0, 0]
[2, 0, 0, 0]
[3, 0, 0, 0]
[4, 0, 0, 0]
[5, 0, 0, 0]

[0, 1, 2, 3]
[1, 1, 2, 3]
[2, 2, 1, 2]
[3, 2, 2, 2]
[4, 3, 3, 2]
[5, 4, 4, 3]
3

Description: Dynamic programming problem sometimes, have to start thinking about thinking from the intermediate state, which is to think how transfer, and then to think about how to define the state.