Leetcode letter combinations of a Phone number (C,c++,java,python)

Problem:

Given a digit string, return all possible letter combinations, the number could represent.

A mapping of Digit to letters (just as on the telephone buttons) is given below.

Input: Digit string "Output": ["Ad", "AE", "AF", "BD", "Be", "BF", "CD", "CE", "CF"].

Note:
Although the above answer is in lexicographical order, your answer could are in any order you want.

Solution: Recursive enumeration can be done.
Title: Given a numeric string, pressing these numbers on the phone keypad requires all possible combinations of strings, each of which may have the following characters: map=["", "" "," abc "," Def "," Ghi "," JKL "," MnO "," PQRS "," TUV "," WXYZ "]

The idea of solving problems: with solution
Java source code (spents 253ms):

public class Solution {char[][] map={"". ToCharArray (), "". ToCharArray (), "ABC". ToCharArray (), "Def". ToCharArray () , "Ghi". ToCharArray (), "JKL". ToCharArray (), "MnO". ToCharArray (), "Pqrs". ToCharArray (), "TUV". ToCharArray (), "WXYZ"    . ToCharArray ()};    int length=0; Public list<string> lettercombinations (String digits) {list<string> res = new Arraylist<string> (        );        Length=digits.length ();        if (length==0) return res;        char[] tmp = new Char[length];        char[] newdigits = Digits.tochararray ();        Getlettercom (RES,0,NEWDIGITS,TMP);    return res;            } private void Getlettercom (list<string> res,int index,char[] digits,char[] tmp) {if (index>=length) {            String letters = new string (TMP);            Res.add (Letters);        Return        } int digit=digits[index]-' 0 ';            for (int i=0;i<map[digit].length;i++) {tmp[index]=map[digit][i]; Getlettercom (Res,index+1,diGITS,TMP); }    }}

C Language Source code (spents 1ms):

void getlettercom (char** res,char* digits,char* tmp,int index,char map[10][5],int *top) {int i,digit=digits[index]-' 0 ';    char* letters;        if (digits[index]==0) {letters= (char*) malloc (sizeof (char) *index);        tmp[index]=0;strcpy (LETTERS,TMP);        res[*top]=letters; (*top) + +;    Return        } for (i=0;map[digit][i];i++) {tmp[index]=map[digit][i];    Getlettercom (Res,digits,tmp,index+1,map,top); }}char** lettercombinations (char* digits, int* returnsize) {char map[10][5]={"", "", "abc", "Def", "Ghi", "JKL", "MnO",     "PQRS", "TUV", "WXYZ"};     char** res,*tmp;     int num=1,length=0,top=0;         while (Digits[length]) {if (digits[length]== ' 0 ' | | digits[length]== ' 1 ') continue;         else if (digits[length]== ' 7 ' | | | digits[length]== ' 9 ') num*=4;         else num*=3;     length++;     } res= (char**) malloc (sizeof (char*) *num);         if (length==0) {*returnsize=0;     return res;     } tmp= (char*) malloc (sizeof (char) *length); Getlettercom (Res,digits,tmp,0,map,&top);     *returnsize=top; return res;}

C + + source code (spents 3ms):

Class Solution {public:    vector<string> lettercombinations (string digits) {        vector<string> res;        Length=digits.size ();        if (length==0) return res;        char* tmp= (char*) malloc (sizeof (char) * (length+1));        Getlettercom (res,0,digits,tmp);        return res;    } Private:    char map[10][5]={"", "" "," abc "," Def "," Ghi "," JKL "," MnO "," PQRS "," TUV "," WXYZ "};    int top=0,length=0;    void getlettercom (vector<string>& res,int index,string digits,char* tmp) {        if (index>=length) {            tmp[index]=0;            String letters=string (TMP);            Res.push_back (Letters);            return;        }        int digit=digits[index]-' 0 ';        for (int i=0;map[digit][i];i++) {            tmp[index]=map[digit][i];            Getlettercom (res,index+1,digits,tmp);}}    ;

Python source code (spents 69ms):

Class Solution:    map=["", "", "abc", "Def", "Ghi", "JKL", "MnO", "PQRS", "TUV", "WXYZ"]    length=0;res=[]    # @par am {string} digits    # @return {string[]}    def lettercombinations (self, digits):        Self.length=len (digits)        self.res=[]        if Self.length==0:return self.res;        Tmp=[' For I in Range (self.length)]        self.getlettercom (0,digits,tmp)        return self.res    def getlettercom ( SELF,INDEX,DIGITS,TMP):        if index>=self.length:            letters= '. Join (TMP)            self.res.append (Letters)            return        Digit=ord (Digits[index])-ord (' 0 ') for        I in range (len (self.map[digit)):            tmp[index]= Self.map[digit][i]            self.getlettercom (index+1,digits,tmp)

Leetcode letter combinations of a Phone number (C,c++,java,python)