Leetcode: Generalized Abbreviation

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

public class Solution {
public List<String> generateAbbreviations(String word) {
List<String> res = new ArrayList<String>();
dfs(0, word.toCharArray(), new StringBuffer(), 0, res);
return res;
}

public void dfs(int pos, char[] word, StringBuffer sb, int count, List<String> res) {
int len = word.length;
int sbOriginSize = sb.length();
if (pos == len) {
if (count > 0) {
sb.append(count);
}
res.add(sb.toString());
}
else {
//choose to abbr word[pos]
dfs(pos+1, word, sb, count+1, res);

//choose not to abbr word[pos]
//first append previous count to sb if count>0
if (count > 0) sb.append(count);
sb.append(word[pos]);
dfs(pos+1, word, sb, 0, res);
}
sb.setLength(sbOriginSize);
}
}