【LeetCode】412. Fizz Buzz

最新推荐文章于 2024-11-23 18:16:46 发布

墨染百城

最新推荐文章于 2024-11-23 18:16:46 发布

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分类专栏： LeetCode 文章标签： leetcode

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问题描述

问题链接：https://leetcode.com/problems/fizz-buzz/#/description

Write a program that outputs the string representation of numbers from 1 to n.

But for multiples of three it should output “Fizz” instead of the number and for the multiples of five output “Buzz”. For numbers which are multiples of both three and five output “FizzBuzz”.

Example:

n = 15,

Return:
[
    "1",
    "2",
    "Fizz",
    "4",
    "Buzz",
    "Fizz",
    "7",
    "8",
    "Fizz",
    "Buzz",
    "11",
    "Fizz",
    "13",
    "14",
    "FizzBuzz"
]

我的代码

思路非常简单，所以直接上代码了。

public class Solution {
    public List<String> fizzBuzz(int n) {
        /*
        思路很简单，从1开始遍历到n，先%3做测试，如果能整除3就把结果+Fizz，
        然后同样的处理5，如果两个都不能整除就结果+数字
        */

        List<String> outStrList = new ArrayList<String>();

        for(int i = 1; i <= n; i++){
            String out = "";
            if(i % 3 == 0){
                out += "Fizz";
            }
            if(i % 5 == 0){
                out += "Buzz";
            }
            if(out.equals("")){
                out += i;
            }
            outStrList.add(out);
        }
        return outStrList;
    }
}

我的代码打败了9.56%的Java代码。来看看讨论区的大神们。

讨论区

Java 4ms solution , Not using “%” operation

链接地址：https://discuss.leetcode.com/topic/63995/java-4ms-solution-not-using-operation

这个的思路非常棒！

public class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> ret = new ArrayList<String>(n);
        for(int i=1,fizz=0,buzz=0;i<=n ;i++){
            fizz++;
            buzz++;
            if(fizz==3 && buzz==5){
                ret.add("FizzBuzz");
                fizz=0;
                buzz=0;
            }else if(fizz==3){
                ret.add("Fizz");
                fizz=0;
            }else if(buzz==5){
                ret.add("Buzz");
                buzz=0;
            }else{
                ret.add(String.valueOf(i));
            }
        } 
        return ret;
    }
}