Elimination Game——LeetCode进阶路

原题链接https://leetcode.com/problems/elimination-game/

题目描述

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

思路分析

给定数字n,对应1~n的数组,从左到右,从第一个数字开始,间隔删除;然后对得到的数组从右到左,从第一个数字开始,间隔删除;按此规律依次迭代。
从左到右,删除1 3 5 ……
从右到左,删除……52 32 1*
即删除元素的规律是一样的,不过一正一反,但可以化为同向。
问题的相似性很明显,而当n=1时,易得剩下的是1,出口get

源码实现

class Solution {
    public int lastRemaining(int n) {
       return n==1?1:2*(n/2+1-lastRemaining(n/2));
    }
}

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转载自blog.csdn.net/Moliay/article/details/88370755