标签：style blog http color os io for art
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
编程之美上的原题，但是书中给出的答案是递归实现，但是不知道给出答案的原作者难道没有测试吗？
很小的case都会超时，例如："trinitrophenylmethylnitramine", "dinitrophenylhydrazine"
递归版：
1 public class Solution { 2 public int minDistance(String word1, String word2) { 3 if(word1 == null || word1.length() == 0) return word2.length(); 4 if(word2 == null || word2.length() == 0) return word1.length(); 5 if(word1.length() > word2.length()) return minDistance(word2,word1);//assume word1 is shorter than 2 6 if(word1.charAt(0) == word2.charAt(0)){ 7 return minDistance(word1.substring(1),word2.substring(1)); 8 }else { 9 int delete = minDistance(word1,word2.substring(1)) + 1; 10 int change = minDistance(word1.substring(1),word2.substring(1)) + 1; 11 return Math.min(delete,change); 12 } 13 } 14 }
这道题在wiki百科中有比较详细的讲解。具体实现用的DP：
DP算法：
维护一个二维矩阵来记录distance的状态：
dinstance[i][j]分别表示字符串word1[0~i]与word2[0~j]的距离
这里需要将distance开到[word1.length() +1][word2.length() + 1]其中[0][0]表示二者都为空串时，distance显然为0.
当i = 0时，distance[0][j] = j （其中 1 <= j <= word2.length()），同理
当j = 0时，distance[i][0] = i （其中 1 <= i <= word1.length()）而distance[i][j]有两种情况
当word1.charAt(i) == word2.charAt(j)时，
显然distance[i][j] = distance[i-1][j - 1];
当word1.charAt(i) != word2.charAt(j)时，
需要考察distance[i - 1][j - 1]、 distance[i][j - 1]、distance[i - 1][j]分别对应了三种情况：修改word1[i] 为word2[j]、删除word2[j]、删除word1[i]，找到这三者中最小的一个数 ，然后+ 1（表示删除操作或者修改操作）
代码如下：
1 public class Solution { 2 public int minDistance(String word1, String word2) { 3 if(word1 == null || word1.length() == 0) return word2.length(); 4 if(word2 == null || word2.length() == 0) return word1.length(); 5 if(word1.length() > word2.length()) return minDistance(word2,word1);//assume word1 is shorter than 2 6 int height = word1.length() + 1,width = word2.length() + 1; 7 int[][] dp = new int[height][width]; 8 for(int i = 0; i < width;i++){ 9 if(i < height){ 10 dp[i][0] = i; 11 } 12 dp[0][i] = i; 13 } 14 for(int i = 1; i < height ; i++){ 15 for(int j = 1; j < width ; j++){ 16 if(word1.charAt(i - 1) == word2.charAt(j - 1)){ 17 dp[i][j] = dp[i - 1][j - 1]; 18 }else{ 19 dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1; 20 } 21 } 22 } 23 return dp[word1.length()][word2.length()]; 24 } 25 }
[leetcode]Edit distance,布布扣,bubuko.com
标签：style blog http color os io for art
原文地址：http://www.cnblogs.com/huntfor/p/3885944.html
