[LeetCode] 17. Letter Combinations of a Phone Number, leetcode17.letter

[LeetCode] 17. Letter Combinations of a Phone Number, leetcode17.letter
[Question]

Given a digit string, return all possible letter combinations that the number coshould 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 cocould be in any order you want.

[Analysis]

None

[Code]

/********************************** Date: * Author: SJF0115 * Subject: 17. letter Combinations of a Phone Number * URL: https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/* result: AC * Source: LeetCode * blog: * *********************************/# include <iostream> # include <vector> using namespace std; class Solution {public: vector <string> letterCombinations (string digits) {vector <string> vec; if (digits. length () <= 0) {vec. push_back (digits); return vec;} // if vector <char> number; DFS (vec, digits, number); return vec;} private: void DFS (vector <string> & vec, string digits, vector <char> & number) {string letters [] = {"", "", "abc ", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; // a combination of int curLen = number. size (); if (curLen = digits. length () {string letter; for (int I = 0; I <curLen; ++ I) {letter + = number [I];} // for vec. push_back (letter); return;} // if // number int num = digits [curLen]-'0 '; // The letter length corresponding to the number int len = letters [num]. length (); for (int I = 0; I <len; ++ I) {number. push_back (letters [num] [I]); DFS (vec, digits, number); number. pop_back () ;}// for }}; int main () {Solution solution; string number = "2"; vector <string> result = solution. letterCombinations (number); // output for (int I = 0; I <result. size (); ++ I) {cout <result [I] <endl ;}// for return 0 ;}

Code 2]

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> result;        DFS(digits,0,"",result);        return result;    }private:    void DFS(string digits,int cur,string path,vector<string> &result){        string keyboard[] = {" ","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};        if(cur == digits.length()){            result.push_back(path);            return;        }//if        int len = keyboard[digits[cur] - '0'].length();        for(int i = 0;i < len;++i){            char c = keyboard[digits[cur] - '0'][i];            DFS(digits,cur + 1,path + c,result);        }//for    }};

Note a special situation: