Letter Combinations of a Phone Number 电话号码的字母组合

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例:

输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

说明:
尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。

思路一：采用递归的方法，每次传入当前遍历的字符串curStr和计数器curCount，如果curCount==nums.size()，那把curStr加入res中，否则就继续递归调用自身，注意当前的curStr加入新的字符在调用完递归函数以后要pop一下，否则curStr会不断累加。

参考代码：

class Solution {
public:
string IntToString(int curN) {
	string res;
	if ((curN >= 2 && curN <= 6)) {
		for (int i = (curN - 2) * 3; i < (curN - 2) * 3 + 3; i++) {
			res.push_back((char)('a' + i));
		}
	}
	else if(curN==7){
		for (int i = 0; i < 4; i++) {
			res.push_back((char)('p' + i));
		}
	}
	else if (curN == 8) {
		for (int i = 0; i < 3; i++) {
			res.push_back((char)('t' + i));
		}
	}
	else if (curN == 9) {
		for (int i = 0; i < 4; i++) {
			res.push_back((char)('w' + i));
		}
	}
	return res;
}
void letterCombinationsCore(string &digits, int curIndex, vector<string> &res, string &curStr) {
	if (curIndex == digits.size()) {
		res.push_back(curStr);
		return;
	}
	for (int i = 0; i < IntToString(digits[curIndex] - '0').size(); i++) {
		curStr.push_back(IntToString(digits[curIndex] - '0')[i]);
		letterCombinationsCore(digits, curIndex + 1, res, curStr);
		curStr.pop_back();
	}
}
vector<string> letterCombinations(string digits) {
	vector<string> res;
	if (digits.empty() || digits.size() == 0) return res;
	string curStr;
	letterCombinationsCore(digits, 0, res, curStr);
	return res;
}
};

思路二：采用循环的方法，相当于为了避免爆栈的问题，每次都选出候选的字符，然后对于每个字符都遍历一遍res，加到res中。最后返回res。

比如我们输入“234”，那么每次考察的都是当前res和候选集的字符，每次把候选集的字符加到res中。

Explanation with sample input "234"

Initial state:

• result = {""}

Stage 1 for number "1":

• result has {""}
• candiate is "abc"
• generate three strings "" + "a", ""+"b", ""+"c" and put into tmp,
  tmp = {"a", "b","c"}
• swap result and tmp (swap does not take memory copy)
• Now result has {"a", "b", "c"}

Stage 2 for number "2":

• result has {"a", "b", "c"}
• candidate is "def"
• generate nine strings and put into tmp,
  "a" + "d", "a"+"e", "a"+"f",
  "b" + "d", "b"+"e", "b"+"f",
  "c" + "d", "c"+"e", "c"+"f"
• so tmp has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
• swap result and tmp
• Now result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }

Stage 3 for number "3":

• result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
• candidate is "ghi"
• generate 27 strings and put into tmp,
• add "g" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
• add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
• add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
• so, tmp has
  {"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
  "adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
  "adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }
• swap result and tmp
• Now result has
  {"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
  "adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
  "adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }

Finally, return result.

参考代码：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
	vector<string> res;
	if (digits.empty()) return res;
	vector<string> map = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
	res.push_back("");
	for (int i = 0; i < digits.size(); i++) {
		string newChars = map[digits[i] - '0'];
		vector<string> tmp;
		for (int i = 0; i < newChars.size(); i++) {
			for (int j = 0; j < res.size(); j++) {
				tmp.push_back(res[j] + newChars[i]);
			}
		}
		res.swap(tmp);
	}
	return res;        
    }
};