LeetCode-412-Fizz Buzz

Fizz Buzz

Title description: Write a program to output a string representation of numbers from 1 to n.

1. If n is a multiple of 3, output "Fizz";

2. If n is a multiple of 5, output "Buzz";

   3. If n is a multiple of 3 and 5 at the same time, output "FizzBuzz".

Please refer to LeetCode official website for example description.

Source: LeetCode
Link: https://leetcode-cn.com/problems/fizz-buzz/
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Solution 1: Traverse

• First, if n is equal to 0, an empty List is returned directly.

• Otherwise, first initialize a List as result, and then traverse the numbers from 1 to n to make a judgment. The judgment process is as follows:

  • If the current number is a multiple of 3 and 5 at the same time, add "FizzBuzz" to result ;
  • If the current number is a multiple of 3, add "Fizz" to result ;
  • If the current number is a multiple of 5, add "Buzz" to result ;
  • Otherwise, the current number will be added to result .
• Finally, return result .

import java.util.ArrayList;
import java.util.List;

/**
 * @Author: ck
 * @Date: 2021/9/29 7:59 下午
 */
public class LeetCode_412 {
    public static List<String> fizzBuzz(int n) {
        List<String> result = new ArrayList<>();
        if (n == 0) {
            return result;
        }
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 && i % 5 == 0) {
                // 同时是3和5的倍数，输出 “FizzBuzz”
                result.add("FizzBuzz");
            } else if (i % 3 == 0) {
                // 是3的倍数，输出“Fizz”
                result.add("Fizz");
            } else if (i % 5 == 0) {
                // 是5的倍数，输出“Buzz”
                result.add("Buzz");
            } else {
                result.add(String.valueOf(i));
            }
        }
        return result;
    }

    public static void main(String[] args) {
        for (String str : fizzBuzz(15)) {
            System.out.println(str);
        }
    }
}

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