Bulb Switcher

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.

分析：http://www.programcreek.com/2014/05/leetcode-bulb-switcher-java/

By using some examples we can find out the number of switches for each bulb:

1 -> 1 (1)
2 -> 2 (1 2)
3 -> 2 (1 3)
4 -> 3 (1 2 4)
5 -> 2 (1 5)
6 -> 4 (1 2 3 6)
7 -> 2 (1 7)
8 -> 4 (1 2 4 8)
9 -> 3 (1 3 9)

So only (i*i)th element has odd number of switches and they are on. The problem is now get all the square numbers.

1 public class Solution {
2     public int bulbSwitch(int n) {
3         return (int)Math.sqrt(n);
4     }
5 }

最直接的方法：

 1 public class Solution {
 2     public int bulbSwitch(int n) {
 3         int count = 0;
 4         for (int i = 1; i <= n; i++) {
 5             int numSwitch = helper(i);
 6             if (numSwitch % 2 == 1)
 7                 count++;
 8         }
 9 
10         return count;
11     }
12 
13     public int helper(int kthBulb) {
14         int count = 0;
15         for (int i = 1; i <= kthBulb; i++) {
16             if (kthBulb % i == 0)
17                 numSwitch++;
18         }
19         return numSwitch;
20     }
21 }

 

转载于:https://www.cnblogs.com/beiyeqingteng/p/5728869.html