390. Elimination Game

问题描述

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

题目链接：

---

思路分析

给一个数字，从左到右隔一个删一个，再从右往左隔一个删一个，直到只剩下最后一个元素，求最后剩下的这个元素

从左往右的情况，奇数偶数的情况都是一样的，都是只剩下偶数，所以就是2乘以从右往左的结果。再看从右往左的结果，如果是奇数，那么还是全是偶数的情况；如果是偶数，那么可以看作奇数的情况再减去1的结果。不断递归出结果即可。

代码

class Solution {
public:
    int lastRemaining(int n) {
        return forward(n);
    }

    int forward(int n){
        if(n == 1) return 1;
        if(n == 2) return 2;
            return 2*backward(n/2);
    }

    int backward(int n){
        if(n == 1 || n == 2) return 1;
        if(n & 1 == 1)
            return 2*forward(n/2);
        else
            return 2*forward(n/2) - 1;
    }
};

时间复杂度：$O(logn)$
空间复杂度：$O(1)$

---

反思

评论区的鬼才一行做法，还不是很理解。

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