LeetCode 72. Edit Distance（编辑距离）

最新推荐文章于 2021-01-26 15:33:04 发布

jmspan

最新推荐文章于 2021-01-26 15:33:04 发布

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分类专栏：动态规划距离编辑距离文本字符串比较递归记忆化搜索文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/jmspan/article/details/51473195

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动态规划同时被 3 个专栏收录

63 篇文章

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51 篇文章

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字符串

30 篇文章

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本文详细介绍了计算两个字符串之间编辑距离的三种方法：动态规划的基本实现、动态规划的内存优化版本以及递归加记忆化搜索的方法。每种方法都提供了详细的代码示例。

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原题网址：https://leetcode.com/problems/edit-distance/

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:

a) Insert a character
b) Delete a character
c) Replace a character

方法一：动态规划。

public class Solution {
    public int minDistance(String word1, String word2) {
        
        char[] wa1 = word1.toCharArray();
        char[] wa2 = word2.toCharArray();
        int[][] distance = new int[wa1.length+1][wa2.length+1];
        for(int len2=1; len2<=wa2.length; len2++) distance[0][len2] = len2;
        for(int len1=1; len1<=wa1.length; len1++) {
            distance[len1][0] = len1;
            for(int len2=1; len2<=wa2.length; len2++) {
                if (wa2[len2-1] == wa1[len1-1]) distance[len1][len2] = distance[len1-1][len2-1];
                else distance[len1][len2] = Math.min(distance[len1-1][len2-1], Math.min(distance[len1][len2-1], distance[len1-1][len2])) + 1;
            }
        }
        return distance[wa1.length][wa2.length];
    }
}

方法二：动态规划，优化内存。

public class Solution {
    public int minDistance(String word1, String word2) {
        char[] wa1 = word1.toCharArray();
        char[] wa2 = word2.toCharArray();
        int[] distance = new int[wa2.length+1];
        int[] buf = new int[wa2.length+1];
        for(int len2=0; len2<=wa2.length; len2++) distance[len2] = len2;
        for(int len1=1; len1<=wa1.length; len1++) {
            int[] prev = distance;
            distance = buf;
            distance[0] = len1;
            for(int len2=1; len2<=wa2.length; len2++) {
                if (wa1[len1-1] == wa2[len2-1]) distance[len2] = prev[len2-1];
                else distance[len2] = Math.min(prev[len2-1], Math.min(distance[len2-1], prev[len2])) + 1;
            }
            buf = prev;
        }
        return distance[wa2.length];
    }
}

方法三：递归+记忆化搜索。

public class Solution {
    int[][] dist;
    boolean[][] visit;
    public int minDistance(String word1, String word2) {
        if (visit == null) {
            visit = new boolean[word1.length() + 1][word2.length() + 1];
            dist = new int[word1.length() + 1][word2.length() + 1];
        }
        if (visit[word1.length()][word2.length()]) return dist[word1.length()][word2.length()];
        if (word1.length() == 0) return word2.length();
        if (word2.length() == 0) return word1.length();

        if (word1.charAt(word1.length() - 1) == word2.charAt(word2.length() - 1)) {
            dist[word1.length()][word2.length()] = 
                minDistance(word1.substring(0, word1.length() - 1), word2.substring(0, word2.length() - 1));
        } else {
            dist[word1.length()][word2.length()] = Math.min(
                minDistance(word1.substring(0, word1.length() - 1), word2),
                minDistance(word1, word2.substring(0, word2.length() - 1)));
            dist[word1.length()][word2.length()] = Math.min(
                dist[word1.length()][word2.length()],
                minDistance(word1.substring(0, word1.length() - 1), word2.substring(0, word2.length() - 1)));
            dist[word1.length()][word2.length()]++;
        }
        visit[word1.length()][word2.length()] = true;
        return dist[word1.length()][word2.length()];
    }
}