[LintCode/LeetCode] Edit Distance

最新推荐文章于 2021-06-18 14:21:18 发布

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最新推荐文章于 2021-06-18 14:21:18 发布

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原文链接：https://segmentfault.com/a/1190000004949824

本文介绍了一个经典的字符串处理问题——编辑距离算法。通过该算法可以计算将一个字符串转换为另一个字符串所需的最少操作次数，包括插入、删除和替换字符。文章详细解释了动态规划的解决方案，并给出了具体的实现代码。

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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

构造dp[i][j]数组，i是word1的index+1，j是word2的index+1，dp[i][j]是将i位的word1转换成j位的word2需要的步数。初始化dp[i][0]和dp[0][i]为dp[0][0]到它们各自的距离i，然后两次循环i和j即可。
理解三种操作：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，发现i多了一位，所以i-1-->j才是所求，需要再删除一位才完成i-->j；而replacement则是换掉word1的最后一位，即之前的i-1-->j-1已经完成，如果word1的第i-1位和word2的第j-1位相等，则不需要替换操作，否则要替换一次完成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];
    }
}