LeetCode 319. Bulb Switcher–C++,java,python 1行解法
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大部分题目C++,Python,Java的解法都有。
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 i-th round, you toggle every i bulb. For the n-th round, you only toggle the last bulb. Find how many bulbs are on after n rounds.
Example:
Input: 3
Output: 1
Explanation:
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.
直接返回平方根就可以了。
C++解法:
class Solution {
public:
int bulbSwitch(int n) {
return sqrt(n);
}
};
java解法:
class Solution {
public int bulbSwitch(int n) {
return (int)Math.sqrt(n);
}
}
python解法:
class Solution:
def bulbSwitch(self, n: int) -> int:
return int(math.sqrt(n))

本文提供了一个简洁的解决方案来解决LeetCode319.BulbSwitcher问题,通过数学原理直接返回给定轮数n后的亮灯数量。使用C++、Java和Python三种语言实现了一行代码解决问题的方法。
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