319. 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 i-th round, you toggle every i bulb. For the n-th round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Example:

Input: 3
Output: 1 
Explanation: 
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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/bulb-switcher
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如果一个数字的所有因子是偶数，那么灯是关的，如果是奇数，及时开的。那么，只有平方数，因子才是奇数。那么结果就是这个数小的所有平方数的个数。

class Solution {
    public int bulbSwitch(int n) {
        int res = 1;
        while (res * res <= n) ++res;
        return res - 1;
    }
}

也可以直接取开方。

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