LeetCode 之 Bulb Switcher

最新推荐文章于 2025-04-12 07:15:00 发布

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本文探讨了一种灯泡开关的问题，通过分析不同轮次中灯泡的状态变化，找到了一种高效求解方法。从初始状态开始，经过多次开关操作后，最终确定亮着的灯泡数量。文中提供了一种优化方案，利用数学规律简化了计算过程。

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 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.

这个题最原始想法如下，对每一个回合遍历一次，n个回合后，输出on的个数：

int bulbSwitch(int n) {
        
        vector<int> bulb(n,0);
        int round_b=1;
        for(;round_b<=n;round_b++){
            int i=0;
            while(i<n){
                i=i+round_b;
                if(i>n) break;
                bulb[i-1]==0?bulb[i-1]=1:bulb[i-1]=0;
            }
        }
        int ans=0;
        for(auto x :bulb){
            ans+=x;
        }
        return ans;
    }

很明显会超时，因为穷举法肯定不是出题人想要的，我仔细研究了一下发现了以下规律：输出为0 n: 0 ；输出为1 n: 1,2,3 ；输出为2 n: 4,5,6,7,8 ；输出为3 n: 9,10,11,12,13,14,15。很明显是1,3,5,7...的规律，所以对于一个n,只需要找到n在第几行就行，代码如下：

int bulbSwitch(int n) {
        /*
        vector<int> bulb(n,0);
        int round_b=1;
        for(;round_b<=n;round_b++){
            int i=0;
            while(i<n){
                i=i+round_b;
                if(i>n) break;
                bulb[i-1]==0?bulb[i-1]=1:bulb[i-1]=0;
            }
        }
        int ans=0;
        for(auto x :bulb){
            ans+=x;
        }
        return ans;*/
        return sqrt(n);
    }