319. Bulb Switcher (M)

原创于 2020-03-14 13:07:29 发布 · 118 阅读

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#整数处理 #因数

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本文探讨了BulbSwitcher算法的核心思想，通过对n个初始关闭的灯泡进行n轮操作，每轮切换特定灯泡状态，最终确定亮着的灯泡数量。文章详细分析了操作规律，揭示了完全平方数与灯泡状态之间的联系，提供了一种高效的解决方案。

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Bulb Switcher (M)

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.

题意

对n个灯泡进行n次操作，第i次操作会切换第k*i个灯泡的状态。问n次操作后会有多少灯泡亮着。

思路

	1	2	3	4	5	6	7	8
状态	O	O	O	O	O	O	O	O
状态	O	X	O	X	O	X	O	X
状态	O	X	X	X	O	O	O	X
状态	O	X	X	O	O	O	O	O
状态	O	X	X	O	X	O	O	O
状态	O	X	X	O	X	X	O	O
状态	O	X	X	O	X	X	X	O
状态	O	X	X	O	X	X	X	X
总次数	1	2	2	3	2	4	2	4

通过找规律不能发现，当一个灯泡被操作的次数为奇数时，这个灯泡最终会保持点亮。而第i个灯泡被操作的次数等于i所对应的因数的个数。而一个数的因数总是成对出现的，如 $8 = 1 * 8 = 2 * 4$ ，只有当出现 $4 = 2 * 2$ 这种完全平方数开根的情况，总因数个数才为奇数。所以问题就转化为了求1-n中完全平方数的个数。

代码实现

class Solution {
    public int bulbSwitch(int n) {
        int res = 0;
        for (int i = 1; i * i <= n; i++) {
            res++;
        }
        return res;
    }
}