Bulb Switcher

题目描述：

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 theith round, you toggle every i bulb. For the nth round, you only toggle the last bulb.Find how many bulbs are on aftern 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.

这个题我真的服！

我自己找到的规律是完全平方数的灯泡肯定会亮着，然后这么写：

public int bulbSwitch(int n) {    int result=0;    for(int i=1;i<=n;i++){    if((int)Math.sqrt(i)*(int)Math.sqrt(i)==i)        result+=1;    }    return result;}

但这么写还超时，我服了。我又想着是不是判断完全平方数这里复杂了，然后自己写了个二分法判断完全平方数，md竟然耗时更长。

public int bulbSwitch(int n) {int result=0;for(int i=1;i<=n;i++){    if(isPerfectSquare(i))    result+=1;    }return result;}public boolean isPerfectSquare(int num){return binaryPerfectSquare(num, 1, num);}public boolean binaryPerfectSquare(int num,int left,int right){if(left>right)return false;int mid=(left+right)/2;if(mid*mid==num)return true;if(mid*mid>num)return binaryPerfectSquare(num, left, mid-1);elsereturn binaryPerfectSquare(num, mid+1, right);}

最后换一个角度，求出比自己小的完全平方数的个数就行了！！！

AC代码如下：

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