有点像约瑟夫问题，但是是一轮正向一轮反向的eliminate。我的brute force是先做一个bool array, mark 还没被删的position
[F, F F, F, F..., F]
-->
<---
空间 O(n), 时间 O(n lgn )?
O(LogN)
O(1)?????!!!!
有点像约瑟夫问题，但是是一轮正向一轮反向的eliminate。我的brute force是先做一个bool array, mark 还没被删的position
[F, F F, F, F..., F]
-->
<---
空间 O(n), 时间 O(n lgn )?
O(LogN)
O(1)?????!!!!
有点像约瑟夫问题，但是是一轮正向一轮反向的eliminate。我的brute force是先做一个bool arr...
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