17. Letter Combinations of a Phone Number



题目：Letter Combinations of a Phone Number

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.

递归法，代码.

void getStr(char* digits,int index,char map[10][5],char** ret,char* tmp,int* sub);/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */char** letterCombinations(char* digits, int* returnSize) {     int i=0;    int num=1;    char** ret;    int sub=0;        int length=strlen(digits);    if(length==0)   return "";        char map[10][5]={" "," ","abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};      char* tmp=(char *)malloc(sizeof(char)*length);//暂时保存的字符串        while(i<length)    {        int t=digits[i++]-'0';        if(t==0 || t==1) return "";        else if(t==7 || t==9) num*=4;        else num*=3;    }        ret=(char **)malloc(sizeof(char *)*num); //开辟空间        getStr(digits,0,map,ret,tmp,&sub);    *returnSize=sub;    return ret;}void getStr(char* digits,int index,char map[10][5],char** ret,char* tmp,int* sub){    int i=0;    int len=strlen(digits);    if(index== len)    {        ret[(*sub)]=(char*)malloc(sizeof(char)*(len+1)); //保存数据        for(i=0;i<len;i++)        {            printf("The character is %c ",tmp[i]);            ret[*sub][i]=tmp[i];        }        ret[*sub][i]='\0';        (*sub)++;        return;    }    int number=digits[index]-'0';    //printf("The number is %d %d\n",number,digits[0]);    for(i=0; i<strlen(map[number]);i++)    {        tmp[index]=map[number][i];        getStr(digits,index+1,map,ret,tmp,sub);    }}