lintcode 779. 廣義縮寫dfs

請完成一個能夠給出 wordword 的所有“縮寫”的函數（給出任意一種排列即可）

樣例
例1:

輸入: 
word = "word", 
輸出: 
["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
例2:

輸入:
word = "today"
輸出:
["1o1a1","1o1ay","1o2y","1o3","1od1y","1od2","1oda1","1oday","2d1y","2d2","2da1","2day","3a1","3ay","4y","5","t1d1y","t1d2","t1da1","t1day","t2a1","t2ay","t3y","t4","to1a1","to1ay","to2y","to3","tod1y","tod2","toda1","today"]

class Solution {
public:
    /**
     * @param word: the given word
     * @return: the generalized abbreviations of a word
     */
    vector<string> generateAbbreviations(string &word) {
        // Write your code here
        vector<string>res;
        string tmp="";
        dfs(word,res,0,tmp,0);
        return res;
    }
    void dfs(string &word,vector<string>&res,int index,string tmp,int cnt)
    {
        if(index==word.size()){
            if(cnt>0) tmp+=to_string(cnt);
            res.push_back(tmp);
            return;
        }
        dfs(word,res,index+1,tmp,cnt+1);
        if(cnt>0) tmp+=to_string(cnt)+word[index];
        else tmp+=word[index];
        dfs(word,res,index+1,tmp,0);
    }
};