LeetCode 319 -Bulb Switcher ( JAVA )

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.

Example:

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.

public class Solution {    public int bulbSwitch(int n) {        return (int)Math.sqrt(n);    }}

总结：这感觉跟编程题目有点不着边了，值得注意的一点是every second bulb，这句话是翻译是，每两个灯的意思， 例如

first ：on on on on ，

second：on off on off，

third： on off off off，

forth： on off off on，

原谅我没仔细研究内部规律，参考是discuss里面的代码；

参考链接：https://leetcode.com/discuss/91371/share-my-o-1-solution-with-explanation