LeetCode题解：Bulb Switcher

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.

题意：有[1,n]个灯泡处于关闭状态，判断到第n轮时有多少个灯泡是亮着的。第一轮，将所有灯泡点亮；第二轮，将所有序号为2的倍数的灯泡点亮；第三轮，将所有序号为3的倍数的灯泡点亮；以此类推

解题思路：灯泡在经过n轮的折腾后，处于亮的状态的话，表示该灯泡的按钮被按了奇数次，例如第i个灯泡，如果其按钮要被按，那么i肯定是[1,n]中某个数的分解因子，对于一个数它的分解因子一般成对出现，例如12：1-12，2-6，3-4，除非该数可以开根号，如36。对于第一种情况，灯泡必然被按偶数次，第二种情况，为奇数次。所以对n开根号就可以知道有多少个数被按了奇数次。

代码：

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