390. Elimination Game

本文探讨了一个有趣的算法问题,从1到n的有序整数列表开始,通过交替从左到右和从右到左移除特定位置的数字,直到只剩下一个数字。通过递归方式解决此问题,并提供了完整的解析思路与代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.

Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.

We keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Find the last number that remains starting with a list of length n.

Example:

Input:
n = 9,
1 2 3 4 5 6 7 8 9
2 4 6 8
2 6
6

Output:
6

=========================================================================

Resolve ideas:

We record two operations as f(x) and g(x), respectively.

When we do the first operation,we can get such a rule. 

  • f(n) = 2*g(n-1)

In the next operation, we draw such a conclusion through mathematical induction.

  • g(n) = -f(n)+n+1

Obviously, these two operations are done alternately.

So we solve it recursively.

==========================================================================

Code:

class Solution {
    public int lastRemaining(int n) {
        return helper(n,true);
    }
    private int helper(int n,boolean flag){
        if (n==1) return 1;
        if (flag)
            return 2*helper(n/2,false);
        else
            return n+1-helper(n,true);
    }
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值