(java)Bulb Switcher

最新推荐文章于 2018-10-21 15:30:48 发布

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最新推荐文章于 2018-10-21 15:30:48 发布

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文章标签： java 编程

本文链接：https://blog.youkuaiyun.com/chaiqunxing51/article/details/51426039

本文探讨了一个经典的灯泡开关问题，通过逐步优化算法从因子计数到寻找完全平方数的方法，最终实现高效的解决方案。

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题意：（开灯关灯的问题，轮训）

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 ith round, you toggle every i bulb. 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.

（1）最后整理的思路：先N以内所有数的因子有多少个，如果是奇数个就count++；

 public static int bulbSwitch(int n) {
		 int[] temp=new int[n+1];
		 for(int i=1;i<=n;i++){
			 for(int j=2;i*j<=n;j++){
				 temp[i*j]++;
			 }
		 }
		 int count=0;
		 for(int i=1;i<=n;i++){
			 if(temp[i]%2==0){
				 count++;
			 }
		 }
		 return count;
	 }

（2）超时，进而思路：N以内只有是能开方的数的因子才是奇数个，所以求N以内能开方（完全平方数比如4,9,16......）的数的个数就行

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

（3）超时，进而思路,，直接求个数（网上更简洁的方式是return Math.sqrt(N)）

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