FizzBuzz

写一个程序，输出从 1 到 n 数字的字符串表示。

1. 如果 n 是3的倍数，输出“Fizz”；

2. 如果 n 是5的倍数，输出“Buzz”；

3.如果 n 同时是3和5的倍数，输出 “FizzBuzz”。

示例：

n = 15,

返回:
[
    "1",
    "2",
    "Fizz",
    "4",
    "Buzz",
    "Fizz",
    "7",
    "8",
    "Fizz",
    "Buzz",
    "11",
    "Fizz",
    "13",
    "14",
    "FizzBuzz"
]

 1 /**
 2  * Return an array of size *returnSize.
 3  * Note: The returned array must be malloced, assume caller calls free().
 4  */
 5 char** fizzBuzz(int n, int* returnSize) {
 6     char  string[] = "0123456789";
 7     int i=0;
 8     char tmp[25]={'\0'};
 9     char ** p;
10     p=(char **)malloc(sizeof(char *) * n); 11 if(!p) return NULL; 12 13 //开辟空间 **p 14 for(i=0;i<n;i++){ 15 p[i] = (char *) malloc(sizeof(char) * 9); 16 if(!p[i]) return NULL; 17 memset(p[i] , '\0',9); 18  } 19 20 for(i=1;i<=n;i++){ 21 if(i%3 == 0 && i%5 == 0) 22 {memcpy(p[i-1], "FizzBuzz",strlen("FizzBuzz")); continue;} 23 else if(i%3 == 0){ 24 memcpy(p[i-1],"Fizz",4); 25 continue; 26  } 27 else if(i%5 == 0){ 28 memcpy(p[i-1],"Buzz",4);continue; 29  } 30 else{ 31 32 //itoa(i,tmp,10); 33 int t=0,j=0,len;int k=i; 34 while(k != 0){ 35 t = k%10; 36 tmp[j++] = string[t]; 37 k = k/10; 38  } 39 //roll back 40 len =j;j=j-1; 41 while(k<len/2){ 42 tmp[k] =tmp[k] ^ tmp[j]; 43 tmp[j] =tmp[j] ^ tmp[k]; 44 tmp[k] =tmp[k] ^ tmp[j]; 45 k++;j--; 46  } 47 memcpy(p[i-1], tmp,strlen(tmp) ); 48  } 49  } 50 //for(i=0;i<n;i++) 51 //printf("%s\n",p[i]); 52 return p; 53 }