390. 消除游戏

arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
arr = [2, 4, 6, 8]
arr = [2, 6]
arr = [6]
//每次都是等差数列
//每次个数n /= 2
//每次d *= 2
//总共四种情况，个数为奇/偶，从左/右开始，只有偶&&从右开始时候a0不变，剩下都是a0 + d

class Solution {
public:
    int lastRemaining(int n) {
        int num_sum = n;
        int a0 = 1;
        int d = 1;
        int left_or_right_begin = 1; //1表示从左开始
        while(num_sum != 1){
            if(left_or_right_begin == -1 && num_sum % 2 == 0){
                a0 = a0;
            }else{
                a0 += d;
            }

            num_sum /= 2;
            d *= 2;
            left_or_right_begin *= -1;
        }
        return a0;
    }
};