Leetcode题目：Letter Combinations of a Phone Number

题目：

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.

题目解答：

题目要求根据输入的数字串，判断出可能产生的字母序列。首先，需要存储每个数字与字母字符的映射关系，在这里，我们使用map结构来存储。之后，不断使用迭代实现对可能产生的字母序列的遍历过程。

代码：

class Solution {
public:
    Solution()  //建立映射关系
    {
        vector<char> _0_;
        digitMapChars['0'] = _0_;
        
        vector<char> _1_;
        digitMapChars['1'] = _1_;
        
        vector<char> _2_;
        _2_.push_back('a'); 
        _2_.push_back('b');
        _2_.push_back('c');
        digitMapChars['2'] = _2_;
        
        vector<char> _3_;
        _3_.push_back('d');
        _3_.push_back('e');
        _3_.push_back('f');
        digitMapChars['3'] = _3_;
        
        vector<char> _4_;
        _4_.push_back('g');
        _4_.push_back('h');
        _4_.push_back('i');
        digitMapChars['4'] = _4_;
        
        vector<char> _5_;
        _5_.push_back('j');
        _5_.push_back('k');
        _5_.push_back('l');
        digitMapChars['5'] = _5_;
        
        vector<char> _6_;
        _6_.push_back('m');
        _6_.push_back('n');
        _6_.push_back('o');
        digitMapChars['6'] = _6_;
        
        vector<char> _7_;
        _7_.push_back('p');
        _7_.push_back('q');
        _7_.push_back('r');
        _7_.push_back('s');
        digitMapChars['7'] = _7_;
        
        vector<char> _8_;
        _8_.push_back('t');
        _8_.push_back('u');
        _8_.push_back('v');
        digitMapChars['8'] = _8_;
        
        vector<char> _9_;
        _9_.push_back('w');
        _9_.push_back('x');
        _9_.push_back('y');
        _9_.push_back('z');
        digitMapChars['9'] = _9_;
    }
    
    vector<string> letterCombinations(string digits) {
    	vector<string> fres; //最终结果存储在fres中
        if(digits == "")
            return fres;
        string::iterator sit = digits.begin();  //使用sit迭代器来完成对digits这个字符串的访问
        list<string> res;  //结果存储在res中
        
        int former_size = 1;
        while(sit != digits.end())
        {
            if(digitMapChars.find(*sit) == digitMapChars.end())  //出现未知字符
            {
                break;
            }
            int tmp_size = digitMapChars[*sit].size();
            if(res.empty())
            {
                former_size *= digitMapChars[*sit].size();
                for(vector<char>::iterator vit = digitMapChars[*sit].begin(); vit != digitMapChars[*sit].end(); vit++)
                {
                	string t = "";
                    res.push_back(t + *vit);
                }
            }
            else
            {
                int former_num = former_size;
                former_size *= digitMapChars[*sit].size();
                int cur_num = digitMapChars[*sit].size();
                while(former_num != 0)
                {
                    string strtmp = res.front();
                    for(int i = 0;i < cur_num;i++)
                    {
                        res.push_back(strtmp + digitMapChars[*sit][i]);
                    }
                    res.pop_front();
                    former_num--;
                }
            }
            sit++;  //使用sit来实现对字符串中数字的遍历
        }
        
        for(list<string>::iterator sit = res.begin(); sit != res.end(); sit++)
        {
        	//cout << *sit << endl;
            fres.push_back(*sit);
        }
        return fres;
        
    }
private:
    map<char,vector<char> > digitMapChars;
};