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:
题目大意:
解题思路:
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)
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