每日算法之十七：Letter Combinations of a Phone Number

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.

下面给出两种方法，本质上是一样的

class Solution {  
//static declare and definition with local scope  
public:  
    void letterCombinations_aux(int step, string& path, vector<string>& ans, const string& digits)  
    {  
        //pay attention to this kind of statement  
        const static string strT[10] = {"","","abc","def","ghi","jkl","mno","qprs","tuv","wxyz"};  
        if(step == digits.size())  
        {  
            ans.push_back(path);  
            return;  
        }  
        for(int i = 0; i < strT[digits[step]-'0'].size(); ++i)  
        {  
            path.push_back(strT[digits[step]-'0'][i]);  
            letterCombinations_aux(step+1, path, ans, digits);  
            path.pop_back();  
        }  
    }  
    vector<string> letterCombinations(string digits) {  
        // Start typing your C/C++ solution below  
        // DO NOT write int main() function  
        string path;  
        vector<string> ans;  
        int step = 0;  
        letterCombinations_aux(step, path, ans, digits);  
        return ans;  
    }  
};  

第二种：

class Solution {
private:
    map<char, vector<char> > dict;
    vector<string> ret;
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 dep, int maxDep, string &s, string ans)
    {
        if (dep == maxDep)
        {
            ret.push_back(ans);
            return;
        }
        
        for(int i = 0; i < dict[s[dep]].size(); i++)
            dfs(dep + 1, maxDep, s, ans + dict[s[dep]][i]);
    }
    
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        createDict();
        dfs(0, digits.size(), digits, "");
        return ret;
    }
};