如何在Skip lists时间复杂度（最差情况）下将2个给定Skip List（每个有n个密钥）合并为一个O(n)？
只是寻找算法 - 没有特定的实现/语言。
答案 0 :(得分：7)
store the two skip lists in two arrays: A,B
merge the arrays.
repeat until getting into root ('top' is a linked list containing only one element):
for each second entry in the skip list 'top' add a 'tag' (link 'upstairs' of the previous level)
确实是O（n）因为存储和合并是O（n），并且在循环中，你需要迭代：
n+n/2+n/4+n/8+... = sigma(n/2^k) where k in (0,infinity) = 2n