LeetCode 17 - Letter Combinations of a Phone Number

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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.

My Code

class Solution {public:    vector<string> letterCombinations(string digits) {        if (digits.empty())        {            vector<string> empty;            return empty;        }        int size = digits.size();        int resultsSize = 1;        for (int i = 0; i < size; i++)        {            char c = digits[i];            if (c == '0')                continue;            else if (c == '7' || c == '9')                resultsSize *= 4;            else                resultsSize *= 3;        }        vector<string> results(resultsSize, string(size, ' '));        int divisor = resultsSize;        for (int i = 0; i < size; i++)        {            char c = digits[i];            char set[4];            int setSize = 3;            if (c == '0')                continue;            else            {                switch (c)                {                    case '1':                        set[0] = '0';                        set[1] = '_';                        set[2] = '0';                        break;                    case '2':                        set[0] = 'a';                        set[1] = 'b';                        set[2] = 'c';                        break;                    case '3':                        set[0] = 'd';                        set[1] = 'e';                        set[2] = 'f';                        break;                    case '4':                        set[0] = 'g';                        set[1] = 'h';                        set[2] = 'i';                        break;                    case '5':                        set[0] = 'j';                        set[1] = 'k';                        set[2] = 'l';                        break;                    case '6':                        set[0] = 'm';                        set[1] = 'n';                        set[2] = 'o';                        break;                    case '7':                        set[0] = 'p';                        set[1] = 'q';                        set[2] = 'r';                        set[3] = 's';                        setSize = 4;                        break;                    case '8':                        set[0] = 't';                        set[1] = 'u';                        set[2] = 'v';                        break;                    case '9':                        set[0] = 'w';                        set[1] = 'x';                        set[2] = 'y';                        set[3] = 'z';                        setSize = 4;                        break;                        break;                    default:                        break;                }            }            divisor /= setSize;            for (int j = 0; j < resultsSize; j++)            {                results[j][i] = set[(j/divisor)%setSize];            }        }        return results;    }};

Runtime: 0 ms

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