Comments (5)
假如随机生成的level = i,那新的node就是一个Level i node,它有i个forward pointers。head有所有的level的forward pointers,所以从Level i开始找,然后把所有最下面i层的previous nodes都收集起来,放在update里。
from algo.
加入level0层当前有1万个数据,而随机生成的level = 1时,这样写的代码的时间复杂度不就是level0层单链表的遍历了吗?应该从当前顶层levelCount -1开始遍历,才能发挥跳表的作用吧。不知道我理解的对不对。
from algo.
那倒是,不过反正它也不是每次都会等于1,就是个概率意义上的复杂度
from algo.
这是个二项分布,主要集中在 half of max level 附近。所以不是问题。
from algo.
我没有互联网经验,不清楚对性能来说有多大差距?算法不就是降低最坏的情况发生概率。修改一个数就能解决这个概率问题。不清楚大家为啥说不是问题。
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