Problema
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.
Vincoli
0 ≤ limit ≤ 10,000. A limit of 0 produces an empty list.
Esempi
Input:
limit = 5
Output atteso:
expected = ["1", "2", "Fizz", "4", "Buzz"]Input:
long long limit = 5;
Output atteso:
std::vector<std::string> expected = {"1", "2", "Fizz", "4", "Buzz"};Input:
long limit = 5;
Output atteso:
string[] expected = new string[] {"1", "2", "Fizz", "4", "Buzz"};Spiegazione
3 is divisible by 3 and 5 is divisible by 5. Other entries keep their number.
Input:
limit = 0
Output atteso:
expected = []Input:
long long limit = 0;
Output atteso:
std::vector<std::string> expected = {};Input:
long limit = 0;
Output atteso:
string[] expected = new string[] {};Spiegazione
There are no integers from 1 through 0, so the list is empty.
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
Each position depends only on the current integer’s divisibility by 3 and 5.
Suggerimento 2
Build the result list in order by checking both divisibility conditions for every number.
Suggerimento 3
Handle the lower bound carefully: if the limit is 0, the answer is an empty list.
Complessità attesa
O(n) time, O(n) extra space
Riferimenti
Questa sfida è adattata per CodeUtility. Segui la traccia e i vincoli sopra.