Letter Combinations of a Phone Number(带for循环的DFS，递归总结)

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.

又加深了对DFS的认识，在递归函数中，需要把握几个点，总能各个击破。

1.就是变化的量要作为参数来传递，这样也就是每个函数在自己的栈中都有该局部变量，这样就可以在回溯到某个点的时候，他的局部变量不会消失。

2.一般的终止条件中，计算结果

代码：

private:
    map<char, vector<char> > dict;
    vector<string> ret;
    int len;
public:
    void createDict()
    {
        dict.clear();
        dict[‘2‘].push_back(‘a‘);
        dict[‘2‘].push_back(‘b‘);
        dict[‘2‘].push_back(‘c‘);
        dict[‘3‘].push_back(‘d‘);
        dict[‘3‘].push_back(‘e‘);
        dict[‘3‘].push_back(‘f‘);
        dict[‘4‘].push_back(‘g‘);
        dict[‘4‘].push_back(‘h‘);
        dict[‘4‘].push_back(‘i‘);
        dict[‘5‘].push_back(‘j‘);
        dict[‘5‘].push_back(‘k‘);
        dict[‘5‘].push_back(‘l‘);
        dict[‘6‘].push_back(‘m‘);
        dict[‘6‘].push_back(‘n‘);
        dict[‘6‘].push_back(‘o‘);
        dict[‘7‘].push_back(‘p‘);
        dict[‘7‘].push_back(‘q‘);
        dict[‘7‘].push_back(‘r‘);
        dict[‘7‘].push_back(‘s‘);
        dict[‘8‘].push_back(‘t‘);
        dict[‘8‘].push_back(‘u‘);
        dict[‘8‘].push_back(‘v‘);
        dict[‘9‘].push_back(‘w‘);
        dict[‘9‘].push_back(‘x‘);
        dict[‘9‘].push_back(‘y‘);
        dict[‘9‘].push_back(‘z‘);
    }
    void dfs(int loc,string digits,string temp)
    {
        if(loc==len){
            ret.push_back(temp);
            //temp.erase(temp.size()-1);在这里擦除是没有用的，这已经是下一个递归函数，回溯到上一个的时候，它的局部变量temp还是三个字母
            return;
        }
        for (int i=0;i<dict[digits[loc]].size();++i)
        {
            temp.push_back(dict[digits[loc]][i]);
            dfs(loc+1,digits,temp);
            temp.erase(temp.size()-1);
        }
    }
    vector<string> letterCombinations(string digits) {
        createDict();
        vector<string> tempres;
        tempres.push_back("");
        if(digits.empty()) return tempres;
        len=digits.size();
        dfs(0,digits,"");
        return ret;
    }
};
int main()
{
    //freopen("C:\\Users\\Administrator\\Desktop\\a.txt","r",stdin);
    Solution so;
    so.letterCombinations("258");
    return 0;
}