LeetCode 390. Elimination Game

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

题目：

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

题解：

Only record the head, while n > 1 keep loop.

When n == 1, return head.

We need to move head in 2 conditions. One is now it is left to right, the other is now it is right to left and remaining n is odd.

We need to move head by step. Each time step *= 2 as n /= 2.

Time Complexity: O(logn).

Space: O(1).

AC Java:

 1 class Solution {
 2     public int lastRemaining(int n) {
 3         boolean lr = true;
 4         int head = 1;
 5         int step = 1;
 6         while(n > 1){
 7             if(lr || n % 2 == 1){
 8                 head += step;
 9             }
10             
11             n /= 2;
12             step *= 2;
13             lr = !lr;
14         }
15         
16         return head;
17     }
18 }

LeetCode 390. Elimination Game

原文：https://www.cnblogs.com/Dylan-Java-NYC/p/12324682.html