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:

1 2	Input: n = 3 Output: ["1","2","Fizz"]

Example 2:

1 2	Input: n = 5 Output: ["1","2","Fizz","4","Buzz"]

Example 3:

1 2	Input: n = 15 Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]

這題不難，但很有意思。

自己的做法

TC: O(n), SC: O(n)  
  class Solution {
    public static List fizzBuzz(int n) {
      List ans = new ArrayList();
      for (int i = 1; i <= n; i++) {
        if (i % 15 == 0) {
          ans.add("FizzBuzz");
        } else if (i % 5 == 0) {
          ans.add("Buzz");
        } else if (i % 3 == 0) {
          ans.add("Fizz");
        } else {
          ans.add(String.valueOf(i));
        }
      }
      return ans;
    }
  }

有意思的做法

TC: O(n), SC: O(n)  
class Solution {
    public List fizzBuzz(int n) {
        List ans = new ArrayList<>();
        for(int i = 1; i <= n; i++){
            ans.add(
                i % 15 == 0 ? "FizzBuzz" :
                i % 5 == 0  ? "Buzz" :
                i % 3 == 0  ? "Fizz" :
                String.valueOf(i)
            );
        }
        return ans;
    }
}

第一次看到三元運算子疊加的寫法，簡潔優雅，謝謝指教。

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Lunski's Clutter

412. Fizz Buzz

自己的做法

有意思的做法