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.

采用两种方式都可以实现DFS和BFS

DFS如下：

class Solution {
public:
  void DFS(int pos, std::string digits, std::string &temp, std::vector<std::string> map, std::vector<std::string> &ans)
{
	if (pos == digits.length())
	{
		ans.push_back(temp);
		return ;
	}
	for (int i = 0; i < map[digits[pos] - ‘0‘].length(); i++)
	{
		temp[pos] = map[digits[pos] - ‘0‘][i];
		DFS(pos + 1, digits, temp, map, ans);
	}
}



std::vector<std::string> letterCombinations(std::string digits)
{
	std::string temp;
	std::vector<std::string> ans;
	temp.resize(digits.length());
	std::string m[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
	std::vector<std::string> map;
	for (int i = 0; i < 10; i++)
	{
		map.push_back(m[i]);
	}
	DFS(0, digits, temp, map, ans);
	return ans;
}
};

BFS如下：

class Solution {
public:
  void DFS(int pos, std::string digits, std::string &temp, std::vector<std::string> map, std::vector<std::string> &ans)
{
	if (pos == digits.length())
	{
		ans.push_back(temp);
		return ;
	}
	for (int i = 0; i < map[digits[pos] - ‘0‘].length(); i++)
	{
		temp[pos] = map[digits[pos] - ‘0‘][i];
		DFS(pos + 1, digits, temp, map, ans);
	}
}



std::vector<std::string> letterCombinations(std::string digits)
{
	std::string temp;
	std::vector<std::string> ans;
	temp.resize(digits.length());
	std::string m[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
	std::vector<std::string> map;
	for (int i = 0; i < 10; i++)
	{
		map.push_back(m[i]);
	}
	DFS(0, digits, temp, map, ans);
	return ans;
}
};