138. Copy List with Random Pointer

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.

摘自：

https://leetcode.com/discuss/44347/java-o-n-solution

https://discuss.leetcode.com/topic/7594/a-solution-with-constant-space-complexity-o-1-and-linear-time-complexity-o-n/2

注意这里节点值不会唯一了，和节点值一一映射的关系就不存在了。

public RandomListNode copyRandomList(RandomListNode head) {
  if (head == null) return null;

  Map<RandomListNode, RandomListNode> map = new HashMap<RandomListNode, RandomListNode>();

  // loop 1. copy all the nodes
  RandomListNode node = head;
  while (node != null) {
    map.put(node, new RandomListNode(node.label));
    node = node.next;
  }

  // loop 2. assign next and random pointers
  node = head;
  while (node != null) {
    map.get(node).next = map.get(node.next);
    map.get(node).random = map.get(node.random);
    node = node.next;
  }

  return map.get(head);
}

O(1) extra space version

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:

public RandomListNode copyRandomList(RandomListNode head) {
	RandomListNode iter = head, next;

	// First round: make copy of each node,
	// and link them together side-by-side in a single list.
	while (iter != null) {
		next = iter.next;

		RandomListNode copy = new RandomListNode(iter.label);
		iter.next = copy;
		copy.next = next;

		iter = next;
	}

	// Second round: assign random pointers for the copy nodes.
	iter = head;
	while (iter != null) {
		if (iter.random != null) {
			iter.next.random = iter.random.next;
		}
		iter = iter.next.next;
	}

	// Third round: restore the original list, and extract the copy list.
	iter = head;
	RandomListNode pseudoHead = new RandomListNode(0);
	RandomListNode copy, copyIter = pseudoHead;

	while (iter != null) {
		next = iter.next.next;

		// extract the copy
		copy = iter.next;
		copyIter.next = copy;
		copyIter = copy;

		// restore the original list
		iter.next = next;

		iter = next;
	}

	return pseudoHead.next;
}