LeetCode 17 Letter Combinations of a Phone Number(C,C++,Java,Python)

Problem:

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 "23"
Output: ["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.

Solution:

递归枚举即可。

题目大意：

给定一个数字字符串，在手机按键上按下这些数字，要求得出所有可能的字符串组合，每个字可能会有的字符有以下这些：
map=[" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]

解题思路：

同Solution

Java源代码（用时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语言源代码（用时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++源代码（用时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源代码（用时69ms）：

class Solution:
map=[" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
length=0;res=[]
# @param {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)