390. Elimination Game

最新推荐文章于 2020-12-09 19:40:19 发布

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最新推荐文章于 2020-12-09 19:40:19 发布

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分类专栏： LeetCode

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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。

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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);
    }
}