LeetCode 17 Letter Combinations of a Phone Number (Phone Number combination)

LeetCode 17 Letter Combinations of a Phone Number (Phone Number combination)
Translation

Returns a combination of letters that can be expressed by a given number string. Shows the ing of a number to a letter (just like a telephone button. Input: numeric string "23" output: [ad, AE, af, bd, be, bf, cd, ce, cf] remarks: although the above answers are unordered, if you want to, your answers can be ordered.

Source image

Original

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 23Output: [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.

Code

It looks like I still use C # To get it in one breath ...... It mainly uses recursion because I don't know Digits Cannot Write countless Foreach To judge.

public class Solution{    IList
  
    list = new List
   
    ();    public Solution()    {        list.Insert(0, abc);        list.Insert(1, def);        list.Insert(2, ghi);        list.Insert(3, jkl);        list.Insert(4, mno);        list.Insert(5, pqrs);        list.Insert(6, tuv);        list.Insert(7, wxyz);    }    public IList
    
      LetterCombinations(string digits)    {        IList
     
       result = new List
      
       (); if (digits.Length == 0) return result; if (digits.Length == 1) { foreach (var a in list.ElementAt(int.Parse(digits[0].ToString()) - 2)) { result.Insert(0, a.ToString()); } } int count = 0; IList
       
         temp = LetterCombinations(digits.Substring(1, digits.Length - 1)); foreach (var a in list.ElementAt(int.Parse(digits[0].ToString()) - 2)) { foreach (var rest in temp) { result.Insert(count++, a.ToString() + rest); } } return result; }}
       
      
     
    
   
  

The following is a copy of C ++ code. Frankly speaking, I cannot write such C ++ code ......

class Solution {public:    vector
  
    letterCombinations(string digits) {                vector
   
     ans;        if(digits.size() == 0)            return ans;        int depth = digits.size();        string tmp(depth, 0);        dfs(tmp, 0, depth, ans, digits);        return ans;    }    void dfs(string &tmp, int curdep, int depth, vector
    
      &ans, string &digits){        if(curdep >= depth){            ans.push_back(tmp);            return ;        }        for(int i = 0; i < dic[digits[curdep] - '0'].size(); ++ i){            tmp[curdep] = dic[digits[curdep] - '0'][i];            dfs(tmp, curdep + 1, depth, ans, digits);        }        return ;    }private:string dic[10] = {{},{},{abc},{def},{ghi},{jkl},{mno},{pqrs},{tuv},{wxyz}};};