[LeetCode]319. Bulb Switcher

Medium

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.

非平方数的因子是成对出现的，如6：1*6，2*3，3*2，6*1一共4对，所以非平方数的因子个数是偶数，toggle偶数次，最终是off。
平方数的因子对数是奇数，如4：1*4，2*2，4*1一共3对，toggle奇数次，最终是on。
因此只需找出n以内有多少个平方数即可。
平方数个数的求法：小于n的平方数的个数，为n的平方根取整

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

参考
204. Count Primes