leetcode 319. Bulb Switcher

最新推荐文章于 2022-02-16 21:26:34 发布

Decorator2015

最新推荐文章于 2022-02-16 21:26:34 发布

阅读量320

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/Decorator2015/article/details/51481778

Leetcode 专栏收录该内容

48 篇文章

订阅专栏

本文探讨了一个有趣的灯泡切换问题，初始所有灯泡关闭，在经过多次特定规则的开关操作后，找出最终保持开启状态的灯泡数量。文章通过示例说明了如何通过规律发现解决问题的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目

There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Given n = 3.
At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off].
So you should return 1, because there is only one bulb is on.

解

public class Solution {     
  public int bulbSwitch(int n) {

    int ret=0;
    for(int i=1;i*i<=n;i++){//根据规律，只有操作奇数次的灯，才会亮，因数分解
        ret++;
    }

    return ret;

    }
}