Letter c语言-Combinations of a Phone Number
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Title:
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.
这道题目题意是将每个数字所代表的字符串中的每个字符与其他数字代表的字符串的每个字符相组合,存入一个新字符串中。
该题是典型的递归算法题,基本操作就是一个,将字符串的每个字符存入新字符串中。然后循环其他数字的字符串,进行同样操作,存入上一个字符串的新的字符串。
Solutions:
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; }
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); } }
核心算法:for循环第一个数字代表的字符串的所有字符,然后对于每一个字符,存储到tmp[index]中。这个index就是存储的序号,比如23,代表的abc和def。index首先为0,表示把a存入tmp[0],然后进入getLetterCom中,在信函数中,index=index+1,也就是1,也就是把d存入tmp[1]中,然后进入getLetterCom中,依次循环。当diguts[index]=0,也就是abc到最后的空字符时,结束。将tmp数据copy到letters中,然后存入res中。
接着进入一个新的循环,index仍然为0,也就是将b存入tmp[0]中,覆盖之前存入的a。然后循环。
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