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);
    }
}