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 1:
REF:
http://www.cnblogs.com/etcow/archive/2012/08/30/2662985.html
http://www.cnblogs.com/etcow/archive/2012/08/30/2662985.html
http://blog.csdn.net/fightforyourdream/article/details/13169573
http://www.cnblogs.com/TenosDoIt/p/3465316.html
相当经典的一道递归题目，而且难度级别为4.
例子： "ababd" -> "ccabab"
先初始化matrix如下。意思是，比如"_" -> "cca" = 2 操作是插入'c','c','a'，共3步。 "abab" -> "+ "_" 删除'a','b','a','b'，共4 步。
| _ | a | b | a | b | d | |
| _ | 0 | 1 | 2 | 3 | 4 | 5 |
| c | 1 | |||||
| c | 2 | |||||
| a | 3 | |||||
| b | 4 | |||||
| a | 5 | |||||
| b | 6 |
然后按照注释里的方法填满表格，返回最后一个数字（最佳解）
| _ | a | b | a | b | d | |
| _ | 0 | 1 | 2 | 3 | 4 | 5 |
| c | 1 | 1 | 2 | 3 | 4 | 5 |
| c | 2 | 2 | 2 | 3 | 4 | 5 |
| a | 3 | 2 | 3 | 2 | 3 | 4 |
| b | 4 | 3 | 2 | 3 | 2 | 3 |
| a | 5 | 4 | 3 | 2 | 3 | 3 |
| b | 6 | 5 | 4 | 3 | 2 | 3 |
1 public class Solution { 2 public int minDistance(String word1, String word2) { 3 if (word1 == null || word2 == null) { 4 return 0; 5 } 6 7 int len1 = word1.length(); 8 int len2 = word2.length(); 9 10 int[][] D = new int[len1 + 1][len2 + 1]; 11 12 for (int i = 0; i <= len1; i++) { 13 for (int j = 0; j <= len2; j++) { 14 if (i == 0) { 15 D[i][j] = j; 16 } else if (j == 0) { 17 D[i][j] = i; 18 } else { 19 if (word1.charAt(i - 1) == word2.charAt(j - 1)) { 20 D[i][j] = D[i - 1][j - 1]; 21 } else { 22 D[i][j] = Math.min(D[i - 1][j - 1], D[i][j - 1]); 23 D[i][j] = Math.min(D[i][j], D[i - 1][j]); 24 D[i][j]++; 25 } 26 } 27 } 28 } 29 30 return D[len1][len2]; 31 } 32 }