[LeetCode]17. Letter Combinations of a Phone Number（自学留存）

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.

Personal tips: 根据下一个数字对应3或4个英文，将数组扩充到原来的3或4倍，分别加上英文。如“23”，原来2，申请vector<string>大小为3，遍历到‘3’时，数组大小扩大到3*3=9，每3个添加‘d' , 'e', 'f'，代码实现如下：

class Solution {
public:
	Solution()
	{
		numToAlpha.insert(pair<char, vector<char>>('2', { 'a','b','c' }));
		numToAlpha.insert(pair<char, vector<char>>('3', { 'd','e','f' }));
		numToAlpha.insert(pair<char, vector<char>>('4', { 'g','h','i' }));
		numToAlpha.insert(pair<char, vector<char>>('5', { 'j','k','l' }));
		numToAlpha.insert(pair<char, vector<char>>('6', { 'm','n','o' }));
		numToAlpha.insert(pair<char, vector<char>>('7', { 'p','q','r','s' }));
		numToAlpha.insert(pair<char, vector<char>>('8', { 't','u','v' }));
		numToAlpha.insert(pair<char, vector<char>>('9', { 'w','x','y','z' }));
	}
	vector<string> letterCombinations(string digits) {
		vector<string> combinations;
		int i = 0;
		if (digits.empty()) return combinations;
		while (digits[i]!='\0')
		{
			if (digits[i] != '0' && digits[i] != '1') break;
			++i;
		}
		if (digits[i] == '\0') return combinations;
		vector<char>::iterator it = numToAlpha[digits[i]].begin();
		while (it!= numToAlpha[digits[i]].end())
		{
			string a; 
			a.push_back(*it++);
			combinations.push_back(a);
		}
		++i;
		while (digits[i]!='\0')
		{
			if (digits[i] == '0' && digits[i] == '1') { ++i; continue; }
			int count = numToAlpha[digits[i]].size() - 1, lastNum = combinations.size(), flag = (count > 2), l = lastNum;
			while (count>0)
			{
				int start = 0;
				while (start < lastNum) { combinations.push_back(combinations[start++]); }
				--count;
			}
			while (l>0)
			{
				combinations[l - 1].push_back(numToAlpha[digits[i]][0]); 
				combinations[l + lastNum - 1].push_back(numToAlpha[digits[i]][1]);
				combinations[l + 2 * lastNum - 1].push_back(numToAlpha[digits[i]][2]);
				if (flag == 1) combinations[l + 3 * lastNum - 1].push_back(numToAlpha[digits[i]][3]);
				--l;
			}
			++i;
		}
		return combinations;
	}
private:
	map<char, vector<char>> numToAlpha;
};