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 1:
Input: n = 3
Output: ["1","2","Fizz"]
Example 2:
Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]
Example 3:
Input: n = 15
Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
Constraints:
Solutions
Solution 1: Simulation
We iterate through each integer from 1 to n. For each integer, we check whether it is a multiple of both 3 and 5, or just a multiple of 3, or just a multiple of 5. Based on the check result, we add the corresponding string to the answer array.
The time complexity is O(n), where n is the integer given in the problem. Ignoring the space consumption of the answer array, the space complexity is O(1).
PythonJavaC++GoJavaScriptPHP
class Solution:
def fizzBuzz(self, n: int) -> List[str]:
ans = []
for i in range(1, n + 1):
if i % 15 == 0:
ans.append('FizzBuzz')
elif i % 3 == 0:
ans.append('Fizz')
elif i % 5 == 0:
ans.append('Buzz')
else:
ans.append(str(i))
return ans(code-box)
class Solution {
public List<String> fizzBuzz(int n) {
List<String> ans = new ArrayList<>();
for (int i = 1; i <= n; ++i) {
String s = "";
if (i % 3 == 0) {
s += "Fizz";
}
if (i % 5 == 0) {
s += "Buzz";
}
if (s.length() == 0) {
s += i;
}
ans.add(s);
}
return ans;
}
}(code-box)
class Solution {
public:
vector<string> fizzBuzz(int n) {
vector<string> ans;
for (int i = 1; i <= n; ++i) {
string s = "";
if (i % 3 == 0) {
s += "Fizz";
}
if (i % 5 == 0) {
s += "Buzz";
}
if (s.empty()) {
s = to_string(i);
}
ans.push_back(s);
}
return ans;
}
};(code-box)
func fizzBuzz(n int) []string {
ans := make([]string, 0, n)
for i := 1; i < n+1; i++ {
switch {
case i%15 == 0:
ans = append(ans, "FizzBuzz")
case i%3 == 0:
ans = append(ans, "Fizz")
case i%5 == 0:
ans = append(ans, "Buzz")
default:
ans = append(ans, strconv.Itoa(i))
}
}
return ans
}(code-box)
/**
* @param {number} n
* @return {string[]}
*/
var fizzBuzz = function (n) {
const ans = [];
for (let i = 1; i <= n; ++i) {
if (i % 15 === 0) {
ans.push('FizzBuzz');
} else if (i % 3 === 0) {
ans.push('Fizz');
} else if (i % 5 === 0) {
ans.push('Buzz');
} else {
ans.push(`${i}`);
}
}
return ans;
};(code-box)
class Solution {
/**
* @param Integer $n
* @return String[]
*/
function fizzBuzz($n) {
$ans = [];
for ($i = 1; $i <= $n; ++$i) {
$s = '';
if ($i % 3 == 0) {
$s .= 'Fizz';
}
if ($i % 5 == 0) {
$s .= 'Buzz';
}
if (strlen($s) == 0) {
$s .= $i;
}
$ans[] = $s;
}
return $ans;
}
}(code-box)
