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https://leetcode.com/problems/copy-list-with-random-pointer/#/description

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

https://discuss.leetcode.com/topic/5831/2-clean-c-algorithms-without-using-extra-array-hash-table-algorithms-are-explained-step-by-step

2 clean C++ algorithms without using extra array/hash table. Algorithms are explained step by step.

https://discuss.leetcode.com/topic/22194/o-n-time-o-1-space-c

O(n) time O(1) Space C++

https://discuss.leetcode.com/topic/12025/c-simple-recursive-solution

C++ simple recursive solution

https://discuss.leetcode.com/topic/7594/a-solution-with-constant-space-complexity-o-1-and-linear-time-complexity-o-n

A solution with constant space complexity O(1) and linear time complexity O(N)

An intuitive solution is to keep a hash table for each node in the list, via which we just need to iterate the list in 2 rounds respectively to create nodes and assign the values for their random pointers. As a result, the space complexity of this solution is O(N), although with a linear time complexity.

As an optimised solution, we could reduce the space complexity into constant. The idea is to associate the original node with its copy node in a single linked list. In this way, we don’t need extra space to keep track of the new nodes.

The algorithm is composed of the follow three steps which are also 3 iteration rounds.

1. Iterate the original list and duplicate each node. The duplicate of each node follows its original immediately.
2. Iterate the new list and assign the random pointer for each duplicated node.
3. Restore the original list and extract the duplicated nodes.

The algorithm is implemented as follows:

https://discuss.leetcode.com/topic/18086/java-o-n-solution

Java O(n) solution