lintcode 680. 分割字符串 dfs

给一个字符串,你可以选择在一个字符或两个相邻字符之后拆分字符串,使字符串由仅一个字符或两个字符组成,输出所有可能的结果

样例
样例1

输入： "123"
输出： [["1","2","3"],["12","3"],["1","23"]]
样例2

输入： "12345"
输出： [["1","23","45"],["12","3","45"],["12","34","5"],["1","2","3","45"],["1","2","34","5"],["1","23","4","5"],["12","3","4","5"],["1","2","3","4","5"]]

思路：简单的深度搜索，每一个字符要么自成一个字符串进行递归，要么就和下一个字符组成字符串进行递归即可

class Solution {
public:
    /*
     * @param : a string to be split
     * @return: all possible split string array
     */
    vector<vector<string>> splitString(string& s) {
        // write your code here
        vector<vector<string>> res;
        dfs(s,res,vector<string>(),0);
        return res;
    }
    void dfs(string& s, vector<vector<string>> &res,vector<string> tmp,int index)
    {
        if(index>=s.size()){
            res.push_back(tmp);return;
        } 
        else
        {
            string t="";
            t+=s[index];
            tmp.push_back(t);
            dfs(s,res,tmp,index+1);
            tmp.pop_back();
            if(index+1<s.size())
            {
                t+=s[index+1];
                tmp.push_back(t);
                dfs(s,res,tmp,index+2);
            }
        }
    }
};