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[LintCode/LeetCode] Edit Distance

snowell 發(fā)布于2019-08-14 15:29 / 2980人閱讀

摘要：構(gòu)造數(shù)組，是的，是的，是將位的轉(zhuǎn)換成位的需要的步數(shù)。初始化和為到它們各自的距離，然后兩次循環(huán)和即可。

Problem

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character

Example

Given word1 = "mart" and word2 = "karma", return 3.

Note

構(gòu)造dp[i][j]數(shù)組，i是word1的index+1，j是word2的index+1，dp[i][j]是將i位的word1轉(zhuǎn)換成j位的word2需要的步數(shù)。初始化dp[i][0]和dp[0][i]為dp[0][0]到它們各自的距離i，然后兩次循環(huán)i和j即可。
理解三種操作：insertion是完成i-->j-1之后，再插入一位才完成i-->j；deletion是完成i-->j之后，發(fā)現(xiàn)i多了一位，所以i-1-->j才是所求，需要再刪除一位才完成i-->j；而replacement則是換掉word1的最后一位，即之前的i-1-->j-1已經(jīng)完成，如果word1的第i-1位和word2的第j-1位相等，則不需要替換操作，否則要替換一次完成i-->j。

Solution

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