class Solution 
{
	const char* To[10] = { "","", "abc", "def", "ghi",
		                       "jkl", "mno", "pqrs","tuv", "wxyz" };
	//常量字符串初始化string 注意此非定义(缺省值)--实例化对象时初始化
	//string numToStr[10] = { "","", "abc", "def", "ghi",
	//						   "jkl", "mno", "pqrs","tuv", "wxyz" };
public:
	void generate(string digit, int i, vector<string>& Vs, string str)
	{
		if (i == digit.size())  //如果本次i == size 
			                    //表明上次的i已经到达一个字符串的末字符
		{
			Vs.push_back(str);  //转化完毕--保存
			return;             //返回上一次递归
		}

		int n = digit[i] - '0'; //字符变数字
		string s = To[n];       //数字对应的字符串

		for (auto ch : s)
		{
			generate(digit, i + 1, Vs, str + ch);
		}
	}
	vector<string> letterCombinations(string digit) 
	{
		vector<string> Vs;
		string s;

		if (digit.empty())
			return Vs;
		
		//此时传的 Vs s 均为空对象
		// 
		//Vs的类型是vector<string>
	    //Vs用来存储多个已转化完毕的字符串的
		//
		//s的类型是string
		//s用来存储每一次转化完毕的字符串
		generate(digit, 0,Vs, s);
		return Vs;
	}
};