Facebook Phone Interview: Phone Number to Letter Combinations

来源：互联网发布：流星网络电视直播编辑：程序博客网时间：2024/04/24 03:07

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;}

This is very straightforward to do it recursively. However, iterative is more elegant.

Think it in this way:

Suppose we have one digit, the output is straightforward, if we have two digits: 23, the output should be add the second digits' mapping to each first digit's mapping.

We can use a vector to store first digit's mapping, and loop through the mapping to add the second digit's mapping.

The total time complexity is (m ^ n) : m is the mapping digits' size, n is the total length of digits.

// do it iteratively.vector<string> letterCombinations(string digits) {  if(digits.size() == 0) return {};  vector<string> maps{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};  vector<string> res;  res.push_back("");  for(int i = 0; i < digits.size(); ++i) {    vector<string> backUp;    for(int j = 0; j < maps[digits[i] - '0'].size(); ++j) {      for(int k = 0; k < res.size(); ++k) {        backUp.push_back(res[k] + maps[digits[i] - '0'][j]);      }    }    res = backUp;  }  return res;}int main(void) {  vector<string> res = letterCombinations("123");  for(int i = 0; i < res.size(); ++i) {    cout << res[i] << endl;  }}

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