Edit Distance

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本文介绍了一种使用动态规划的方法来解决两个字符串之间的编辑距离问题。通过插入、删除或替换字符的操作，找到从一个字符串转换为另一个字符串所需的最少步骤。

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) {
		// Start typing your Java solution below
		// DO NOT write main() function
		// for all i and j, p[i,j] will hold the Levenshtein distance between
		// the first i characters of s and the first j characters of t;
		// note that p has (m+1)*(n+1) values
		int n1 = word1.length() + 1;
		int n2 = word2.length() + 1;

		int[][] p = new int[n1][n2];
		for (int i = 0; i < n1; ++i)
			p[i][0] = i;
		for (int j = 0; j < n2; ++j)
			p[0][j] = j;
		for (int i = 1; i < n1; ++i) {
			for (int j = 1; j < n2; ++j) {
				p[i][j] = p[i - 1][j - 1];
				if (word1.charAt(i - 1) != word2.charAt(j - 1))
					++p[i][j];// a substitution
				int min;
				if (p[i - 1][j] < p[i][j - 1])
					min = p[i - 1][j] + 1;// a deletion
				else
					min = p[i][j - 1] + 1;// an insertion
				if (min < p[i][j])
					p[i][j] = min;
			}
		}
		return p[n1 - 1][n2 - 1];
	}
}