README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0412.Fizz%20Buzz/README_EN.md	Math String Simulation

412. Fizz Buzz

中文文档

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:

• 1 <= n <= 104

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 $O(1)$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

func fizzBuzz(n int) (ans []string) {
	for i := 1; i <= n; i++ {
		s := &strings.Builder{}
		if i%3 == 0 {
			s.WriteString("Fizz")
		}
		if i%5 == 0 {
			s.WriteString("Buzz")
		}
		if s.Len() == 0 {
			s.WriteString(strconv.Itoa(i))
		}
		ans = append(ans, s.String())
	}
	return
}

JavaScript

/**
 * @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;
};

PHP

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;
    }
}