Edit Distance

题目原型：

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

基本思路：

设start1和start2 为word1和word2的待比较位置，初始为0,0

那么：
如果word1.charAt(start1)==word2.charAt(start2)时，继续求minDistance(word.subString(start1+1,start2+1));
如果不等于则分六种情况
*1、在word1的start1前面插入和word2的start2处的相等字符，然后求1+minDistance(word.subString(start1,start2+1))
* 2、在word2的start2前面插入和word1的start1处的相等字符，然后求1+minDistance(word.subString(start1+1,start2))
* 3、删除word1的start1处的字符，然后求1+minDistance(word.subString(start1+1,start2))
* 4、删除word2的start2处的字符，然后求1+minDistance(word.subString(start1,start2+1))
* 5、更换word1的start1处为word2的start2处的字符，然后求1+minDistance(word.subString(start1+1,start2+1))
* 6、更换word2的start2处为word1的start1处的字符，然后求1+minDistance(word.subString(start1+1,start2+1))
   最后求这六种情况中最小的值

下面是递归做法：（重复步骤多，太耗时）

public int minDistance(String word1, String word2){return getMin(word1, 0, word1.length()-1, word2,0, word2.length()-1);}public int getMin(String word1, int start1,int end1,String word2,int start2,int end2){if(start1>end1){if(start2>end2)return 0;elsereturn end2-start2+1;}if(start2>end2){if(start1>end1)return 0;elsereturn end1-start1+1;}if(word1.charAt(start1)==word2.charAt(start2))return getMin(word1, start1+1, end1, word2, start2+1, end2);else{int one = 0;int two = 0;int three = 0;one = 1+getMin(word1, start1+1, end1, word2, start2+1, end2);two = 1+getMin(word1, start1+1, end1, word2, start2, end2);three = 1+getMin(word1, start1, end1, word2, start2+1, end2);return Math.min(one, Math.min(two, three));}}

回溯

public int minDistance(String word1, String word2){int[][] rs = new int[word1.length()+1][word2.length()+1];//初始化for(int i = word1.length();i>=0;i--)rs[i][word2.length()] = word1.length()-i;for(int i = word2.length();i>=0;i--)rs[word1.length()][i] = word2.length()-i;for(int i = word1.length()-1;i>=0;i--){for(int j = word2.length()-1;j>=0;j--){if(word1.charAt(i)==word2.charAt(j))rs[i][j] = rs[i+1][j+1];elsers[i][j] = Math.min(rs[i+1][j+1], Math.min(rs[i+1][j], rs[i][j+1]))+1;}}return rs[0][0];}