Question
Formatted question description: https://leetcode.ca/all/390.html
390 Elimination Game
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
Algorithm
It is very simple to do it with recursion. Use a bool type variable left2right
. True means to traverse from left to right, and false means to traverse from right to left.
When n is 1, it returns 1 no matter from left to right or from right to left.
If n is greater than 1,
 If it is from left to right, return twice the traversal of n/2 from right to left;
 If it is from right to left, it will be a little more troublesome. It is definitely necessary to call the recursive function on n/2, but it should be divided into parity.
 If n is an odd number, return twice the value of n/2 traversal from left to right;
 If n is an even number, double the traversal value of n/2 from left to right, and subtract 1
Code
Java

public class Elimination_Game { public static void main(String[] args) { Elimination_Game out = new Elimination_Game(); Solution s = out.new Solution(); System.out.println(s.lastRemaining(9)); } class Solution { public int lastRemaining(int n) { return dfs(n, true); } int dfs(int n, boolean left2right) { if (n == 1) return 1; if (left2right) { return 2 * dfs(n / 2, false); } else { return 2 * dfs(n / 2, true)  1 + n % 2; } } } class Solution2 { public int lastRemaining(int n) { return n == 1 ? 1 : 2 * (1 + n / 2  lastRemaining(n / 2)); } } }

Todo

class Solution(object): def lastRemaining(self, n): """ :type n: int :rtype: int """ count = n head = 1 isFromLeft = True step = 1 while count > 1: if isFromLeft or count % 2 == 1: head = head + step count /= 2 step *= 2 isFromLeft = not isFromLeft return head