leetcode Edit Distance

Edit Distance 原题地址：

https://oj.leetcode.com/problems/edit-distance/

The edit distance of two strings, s1 and s2, is defined as the minimum number of point mutations required to change s1 into s2, where a point mutation is one of:

1. change a letter,
2. insert a letter or
3. delete a letter

计算两个字符串s1+ch1, s2+ch2的编辑距离有这样的性质:

1.         d(s1,””) = d(“”,s1) = |s1|    d(“ch1”,”ch2”) = ch1 == ch2 ? 0 : 1;

2.         d(s1+ch1,s2+ch2) = min(d(s1,s2)+ ch1==ch2 ? 0 : 1 , d(s1+ch1,s2), d(s1,s2+ch2) );

  第一个性质是显然的。

              第二个性质:          由于我们定义的三个操作来作为编辑距离的一种衡量方法。

                                          于是对ch1,ch2可能的操作只有

1.         把ch1变成ch2

2.         s1+ch1后删除ch1              d = (1+d(s1,s2+ch2))

3.         s1+ch1后插入ch2              d = (1 + d(s1+ch1,s2))

                                          对于2和3的操作可以等价于:

                                          _2.   s2+ch2后添加ch1              d=(1+d(s1,s2+ch2))

                                          _3.   s2+ch2后删除ch2              d=(1+d(s1+ch1,s2))

                     因此可以得到计算编辑距离的性质2。

public class Solution {    public int minDistance(String word1, String word2) {        int len1 = word1.length();int len2 = word2.length();if (len1 == 0)return len2;if (len2 == 0)return len1;int[][] dist = new int[len1+1][len2+1];for (int i = 0; i <= len1; i++)dist[i][0] = i;for (int j = 0; j <= len2; j++)dist[0][j] = j;for (int i = 1; i <= len1; i++)for (int j = 1; j <= len2; j++) {if (word1.charAt(i-1) == word2.charAt(j-1))dist[i][j] = dist[i-1][j-1];elsedist[i][j] = Math.min(Math.min(dist[i-1][j-1] + 1, dist[i-1][j] + 1), dist[i][j-1] + 1);}return dist[len1][len2];    }}