[LeetCode 319] Bulb switch-优快云博客

本文链接：https://blog.youkuaiyun.com/sbitswc/article/details/50415951

本文探讨了一个有趣的灯泡开关问题，通过分析不同轮次中灯泡的状态变化规律，揭示了最终点亮灯泡数量与数字因数之间的关系。特别地，文章提出了一种高效的算法来求解任意轮次后点亮灯泡的数量。

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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 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.

Solution:

go through some example from 1 to 5, there is a pattern, we need to find numbers of factor of the number.

Like,

2 - 1,2

3 - 1,3

4 - 1,2,4

5 - 1,5

6- 1,2,3,6

9- 1,3,9

when you add the number which has odd number of factors, the result will increase 1. Every number has even number factors, except square number(1,4,9)

Same with this question Sqrt(x)

public int bulbSwitch(int n) {
        long low = 0;
        long high = n;
        if(n<=0) return 0;
        if(n==1) return 1;
        while(high - low >1) {
            long mid = low + (high-low)/2;
            if(mid * mid <=n) {
                low = mid;
            }else {
                high = mid;
            }
        }
        return (int) low;
    }