Facebook Phone Interview: Phone Number to Letter Combinations (easy)

Input: string of digits of arbitrary length

output: print all possible letter combinations for those digits to screen.

// This is a typical DFS algorithm.
// Suppose we dont know the phone screen layout. we need to set a dictionary first.
// 
map<char, vector<char> > dict; // we use a map structure to map digits to letters.
vector<string> ret;
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');
}

// dfs algorithm.
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) {
    ret.clear();
    createDict();
    dfs(0, digits.size(), digits, "");
    return ret;
}