LeetCode-edit-distance

最新推荐文章于 2022-02-06 11:01:06 发布

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最新推荐文章于 2022-02-06 11:01:06 发布

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分类专栏： LeetCode 文章标签： edit-distance LeetCode

本文链接：https://blog.youkuaiyun.com/weixin_42146769/article/details/89054533

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本文介绍了一种计算两个字符串之间编辑距离的算法，即通过插入、删除或替换字符将一个字符串转换为另一个字符串所需的最少操作次数。该算法适用于拼写检查、DNA序列比对等场景，并提供了详细的Java实现代码。

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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

public class Solution {
    public int minDistance(String word1, String word2) {
        if(word1 == null || word1.length() == 0) {
            return word2 == null ? 0 : word2.length();
        }
 
        if(word2 == null || word1.length() == 0)
            return word1.length();
 
        int l1 = word1.length();
        int l2 = word2.length();
 
        int[][] res = new int[l1+1][l2+1];
 
        for(int i = 1; i <= l1; i++)
            res[i][0] = i;
        for(int i = 1; i <= l2; i++)
            res[0][i] = i;
 
        for(int i = 1; i <= l1; i++) {
            for(int j = 1; j <= l2; j++) {
                if(word1.charAt(i-1) == word2.charAt(j-1)) {
                    res[i][j] = Math.min(Math.min(res[i-1][j], res[i][j-1]) + 1, res[i-1][j-1]);
                } else {
                    res[i][j] = Math.min(res[i-1][j-1], Math.min(res[i-1][j], res[i][j-1])) + 1;
                }
            }
        }
 
        return  res[l1][l2];
    }
}