---
title: 'LeetCode 412. Fizz Buzz'
disqus: hackmd
---

# LeetCode 412. Fizz Buzz

## Description

Given an integer n, return a string array answer (1-indexed) where:

* answer[i] == "FizzBuzz" if i is divisible by 3 and 5.
* answer[i] == "Fizz" if i is divisible by 3.
* answer[i] == "Buzz" if i is divisible by 5.
* answer[i] == i (as a string) if none of the above conditions are true.

## Example

Input: n = 3
Output: ["1","2","Fizz"]

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]

## Constraints

1 <= n <= 10^4^

## Answer

此題主要是考雙重指標，取餘數為0給定不同字串。在數字時注意使用 "snprintf" function可以將數字化成字串。

//2022_03_13
void mystrcpy(char *s, char *t);
char ** fizzBuzz(int n, int* returnSize){
    char **ans = (char**)malloc(sizeof(char*)*n);
    *returnSize = 0;
    char *FB = "FizzBuzz", *F = "Fizz", *B = "Buzz";
    for(int i = 1; i <= n; i++){
        if(i%3==0 && i%5==0){
            ans[*returnSize] = (char*)malloc(sizeof(char)*9);
            mystrcpy(ans[*returnSize],FB);
        }
        else if(i%3==0){
            ans[*returnSize] = (char*)malloc(sizeof(char)*5);
            mystrcpy(ans[*returnSize],F);
        }
        else if(i%5==0){
            ans[*returnSize] = (char*)malloc(sizeof(char)*5);
            mystrcpy(ans[*returnSize],B);
        }
        else{
            ans[*returnSize] = (char*)malloc(sizeof(char)*5);
            snprintf(ans[*returnSize],10,"%d",i);
        }
        (*returnSize)++;
    }
    return ans;
}

void mystrcpy(char *s, char *t){
    while(*t != '\0'){
        *s++ = *t++;
    }
    *s = '\0';
}

## Link
https://leetcode.com/problems/fizz-buzz/
###### tags: `Leetcode`