問題
For each integer from 1 through limit, use Fizz for multiples of 3, Buzz for multiples of 5, and FizzBuzz for multiples of both. Otherwise use the number written as text. Return the entries in order instead of printing them.
Implement fizz_buzz(limit).
Input
limit: The last integer to include.
Output
Return a list of strings, one for each integer from 1 through limit.
制約
0 ≤ limit ≤ 10,000. A limit of 0 produces an empty list.
例
入力：
limit = 5
期待される出力：
expected = ["1", "2", "Fizz", "4", "Buzz"]入力：
long long limit = 5;
期待される出力：
std::vector<std::string> expected = {"1", "2", "Fizz", "4", "Buzz"};入力：
long limit = 5;
期待される出力：
string[] expected = new string[] {"1", "2", "Fizz", "4", "Buzz"};解説
3 is divisible by 3 and 5 is divisible by 5. Other entries keep their number.
入力：
limit = 0
期待される出力：
expected = []入力：
long long limit = 0;
期待される出力：
std::vector<std::string> expected = {};入力：
long limit = 0;
期待される出力：
string[] expected = new string[] {};解説
There are no integers from 1 through 0, so the list is empty.
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Each position depends only on the current integer’s divisibility by 3 and 5.
ヒント 2
Build the result list in order by checking both divisibility conditions for every number.
ヒント 3
Handle the lower bound carefully: if the limit is 0, the answer is an empty list.
目安の計算量
O(n) time, O(n) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。