319.Bulb Switcher

tags: Medium,Math

319. Bulb Switcher

題目描述

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 the i^th round, you toggle every i bulb. For the n^th round, you only toggle the last bulb.

Return the number of bulbs that are on after n rounds.

範例

Example 1:

Input: n = 3
Output: 1
Explanation: At first, the three bulbs are [off, off, off].
After the first round, the three bulbs are [on, on, on].
After the second round, the three bulbs are [on, off, on].
After the third round, the three bulbs are [on, off, off]. 
So you should return 1 because there is only one bulb is on.

Example 2:

Input: n = 0
Output: 0

Example 3:

Input: n = 1
Output: 1

Constraints:

• 0 <= n <= 10⁹

解答

C++

#include <iostream>
#include <cmath>

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

};

順神Apr 26, 2023

Javascript

function bulbSwitch(n) {
  return Math.floor(Math.sqrt(n));
}

寫完看到一堆一行超人，只能跟上了
MarsgoatApr 27, 2023

C#

public int BulbSwitch(int n) {
    return (int)Math.Sqrt(n);
}

JimApr 27, 2023

Python

class Solution:
    def bulbSwitch(self, n: int) -> int:
        return int(sqrt(n))

Ron ChenThu, Apr 27, 2023

Reference

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