LeetCode Edit Distance

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本文介绍了一个经典的编辑距离算法实现，该算法能够计算两个字符串之间的最小编辑距离，即通过插入、删除或替换字符的方式将一个字符串转换为另一个字符串所需的最少操作步骤。文章提供了详细的Java代码示例，展示了如何初始化及计算动态规划矩阵。

Description:

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

Solution:

classical edit distance

pay attention to the initialization circumstances, we can get dp[i][0] = i and dp[0][j] = j

<span style="font-size:18px;">import java.util.*;

public class Solution {
	public int minDistance(String word1, String word2) {
		if (word1 == null)
			return 0;
		if (word2 == null)
			return 0;

		int n = word1.length();
		int m = word2.length();
		if (n == 0)
			return m;
		if (m == 0)
			return n;

		int dp[][] = new int[n + 1][m + 1];

		for (int i = 0; i <= n; i++)
			dp[i][0] = i;
		for (int j = 0; j <= m; j++)
			dp[0][j] = j;

		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++) {
				if (word1.charAt(i - 1) == word2.charAt(j - 1))
					dp[i][j] = Math.min(dp[i - 1][j - 1],
							Math.min(dp[i - 1][j], dp[i][j - 1]) + 1);
				else
					dp[i][j] = Math.min(dp[i - 1][j - 1] + 1,
							Math.min(dp[i - 1][j], dp[i][j - 1]) + 1);
			}

		return dp[n][m];
	}
}</span>