Write a function to generate the generalized abbreviations of a word. Example: Given word = "word", return the following list (order does not matter): ["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
这道题肯定是DFS/Backtracking, 但是怎么DFS不好想，跟Leetcode: Remove Invalid Parentheses的backtracking很像。
Generalized Abbreviation这道题是当前这个字母要不要abbreviate，要或者不要两种选择，Parentheses那道题是当前括号要不要keep在StringBuffer里，要或不要同样是两种选择。
Syntax：注意27行使用StringBuffer.setLength(), 因为count一直累加可能变成两位数三位数，delete stringbuffer最后一个字母可能不行，所以干脆设置为最初进recursion的长度
参考了：https://leetcode.com/discuss/76783/easiest-14ms-java-solution-beats-100%25
1 public class Solution { 2 public List<String> generateAbbreviations(String word) { 3 List<String> res = new ArrayList<String>(); 4 dfs(0, word.toCharArray(), new StringBuffer(), 0, res); 5 return res; 6 } 7 8 public void dfs(int pos, char[] word, StringBuffer sb, int count, List<String> res) { 9 int len = word.length; 10 int sbOriginSize = sb.length(); 11 if (pos == len) { 12 if (count > 0) { 13 sb.append(count); 14 } 15 res.add(sb.toString()); 16 } 17 else { 18 //choose to abbr word[pos] 19 dfs(pos+1, word, sb, count+1, res); 20 21 //choose not to abbr word[pos] 22 //first append previous count to sb if count>0 23 if (count > 0) sb.append(count); 24 sb.append(word[pos]); 25 dfs(pos+1, word, sb, 0, res); 26 } 27 sb.setLength(sbOriginSize); 28 } 29 }
Leetcode: Generalized Abbreviation
原文：http://www.cnblogs.com/EdwardLiu/p/5092886.html