给你一个整数 n ，找出从 1 到 n 各个整数的 Fizz Buzz 表示，并用字符串数组 answer（下标从 1 开始）返回结果，其中：
answer[i] == "FizzBuzz"如果i同时是3和5的倍数。answer[i] == "Fizz"如果i是3的倍数。answer[i] == "Buzz"如果i是5的倍数。answer[i] == i（以字符串形式）如果上述条件全不满足。
示例 1：
输入：n = 3 输出：["1","2","Fizz"]
示例 2：
输入：n = 5 输出：["1","2","Fizz","4","Buzz"]
示例 3：
输入：n = 15 输出：["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
leetcode代码
class Solution {
public:vector<string> fizzBuzz(int n) {vector<string> answer(n);for(int i=0;i<n;i++){if((i+1)%3==0&&(i+1)%5==0) answer[i]="FizzBuzz";else if((i+1)%3==0)answer[i]="Fizz";else if((i+1)%5==0) answer[i]="Buzz";else answer[i]=to_string(i+1);}return answer;}
};总的来说，这道题的思路比较简单，就是模拟+字符串，但需要注意的点很多，比如声明动态数组时必须指定大小，否则不能对answer[i]进行赋值，只能使用push_back或者emplace_back添加元素。并且数组下标从零开始，因此我们要对i+1做判断。
