leetcode 17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

class Solution {
	map<int, vector<char>>num2char;
	void init()
	{
		num2char[2].push_back('a');
		num2char[2].push_back('b');
		num2char[2].push_back('c');
		num2char[3].push_back('d');
		num2char[3].push_back('e');
		num2char[3].push_back('f');
		num2char[4].push_back('g');
		num2char[4].push_back('h');
		num2char[4].push_back('i');
		num2char[5].push_back('j');
		num2char[5].push_back('k');
		num2char[5].push_back('l');
		num2char[6].push_back('m');
		num2char[6].push_back('n');
		num2char[6].push_back('o');
		num2char[7].push_back('p');
		num2char[7].push_back('q');
		num2char[7].push_back('r');
		num2char[7].push_back('s');
		num2char[8].push_back('t');
		num2char[8].push_back('u');
		num2char[8].push_back('v');
		num2char[9].push_back('w');
		num2char[9].push_back('x');
		num2char[9].push_back('y');
		num2char[9].push_back('z');
		
	}
	void do_once(vector<string>&re, string digits)
	{
		vector<string>newre;
		for (int i = 0; i < re.size(); i++)
		{
			for (int j = 0; j < num2char[digits[0] - '0'].size(); j++)
			{
				string str = re[i];
				str += num2char[digits[0] - '0'][j];
				newre.push_back(str);
			}
		}
		re = newre;
	}
public:
	vector<string> letterCombinations(string digits) {
		vector<string>re;
		if (digits.empty())
			return re;
		init();
		re.push_back("");
		while (!digits.empty())
		{
			do_once(re, digits);
			digits.erase(0, 1);
		}
		return re;
	}
};

accepted