【leetcode】Letter Combinations of a Phone Number

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

int counts[] = {0, 0, 3, 3, 3, 3, 3, 4, 3, 4};
char letter[] = {'0', '0', 'a', 'd', 'g', 'j', 'm', 'p', 't', 'w'};
        
class Solution {
public:
    void comb(vector<string> &result, string &str, string &digits, int pos, int size) {
        if (pos == size)
        {
            result.push_back(str);      // save results
            return;
        }
        
        int j = digits[pos]-'0';
        for (int i = 0; i < counts[j]; ++i)
        {
            str[pos] = letter[j] + i;
            comb(result, str, digits, pos+1, size);
        }
    }
    
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int nSize = digits.size();
        string str(nSize, ' ');
        vector<string> result;
        
        comb(result, str, digits, 0, nSize);
        return result;
    }
};

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> result;
        
        getRet(digits, "", 0, result);
        return result;
    }
    
    void getRet(string digits, string s, int pos, vector<string>  &result) {
        const static string dic[10]={ "0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        int size = digits.size();
        
        if(pos == size) {
            result.push_back(s);    // save result
        }
        
        int num = digits[pos] - '0';
        for( int i=0; i<dic[num].size(); i++)   // assemble letters
        {
            string tmp = s;
            tmp += dic[num][i];
            getRet(digits, tmp, pos+1, result);
        }
    }
};