[LeetCode 319] Bulb switch

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;
    }