README_EN.md

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

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.size() == 0) s = to_string(i);
            ans.push_back(s);
        }
        return ans;
    }
};

Go

func fizzBuzz(n int) []string {
	var 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 ans
}

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