Copy A Linked List With Next And Random Pointer

最新推荐文章于 2024-09-06 04:39:09 发布

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本文介绍了一种在O(n)时间复杂度内复制带有随机指针的双向链表的方法。首先创建原始链表中每个节点的副本并插入到相应节点之后，接着调整这些副本节点的随机指针指向，最后将原始链表与副本链表分离。

Question: There is a doubly linked list with next pointer and random pointer (points to an arbitrary node in list). You have to make a copy of the linked list and return. In the end original list shouldn't be modified. Time complexity should be O(n).

Modified structure:

class Node {
      int val;
      Node next;
      Node random;
      Node(int x) { val = x; }
  }

Example: In below figure BLUE arrows show next pointer while RED arrows show random pointer.

Answer:

1) Create the copy of every node in the list and insert it in original list between current and next node.

create the copy of A and insert it between A & B..
create the copy of B and insert it between B & C..
Continue in this fashion, add the copy of N to Nth node.

As a result of this exercise, we will get a list like shown below:

2) Now copy the arbitrary link in this fashion

original->next->random = original->random->next;

/*Traverse two nodes in every iteration*/

This works because original->next is nothing but copy of original and Original->random->next is nothing but copy of random.

3) Now restore the original and copy linked lists in this fashion in a single loop.
original->next = original->next->next;
copy->next = copy->next->next;

While doing this, take care of end of list (NULL pointer) and NULL pointer dereference.
So in this manner, we are copying the list in O(n) time and O(1) space complexity.

public class Solution {
	public static void main(String[] args) {
		Node n1 = new Node(1);
		Node n2 = new Node(2);
		Node n3 = new Node(3);
		Node n4 = new Node(4);
		Node n5 = new Node(5);
		
		n1.next = n2;
		n2.next = n3;
		n3.next = n4;
		n4.next = n5;
		
		n1.random = n4;
		n2.random = n1;
		n3.random = n2;
		n4.random = n5;
		n5.random = n3;
		
		Node node = copy(n1);
		while (node != null) {
			System.out.println(node.val);
			System.out.println(node.random.val);
			node = node.next;
		}
	}
	
	public static Node copy(Node listHead) {
		if (listHead == null) return null;
		Node head = listHead;
		Node next = null;
		//Create the copy of every node in the list and insert 
		//it in original list between current and next node.   
		while (head != null) {
			Node n = new Node(head.val);
			next = head.next;
			head.next = n;
			n.next = next;
			head = next;
		}
		Node newListHead = listHead.next;
		//Copy the arbitrary link for result  
		head = listHead;
		while (head != null) {
			head.next.random = head.random.next;
			head = head.next.next;
		}
		
		head = listHead;
		Node temp = head.next;
		//restore the original and copy linked lists 
		while(head != null && temp != null) {
			head.next = temp.next;
			if (temp.next == null) {
				break;
			}
			temp.next = temp.next.next;
			head = head.next;
			temp = temp.next;
		}
		return newListHead;		
	}
}

From: http://tech-queries.blogspot.com/2011/04/copy-linked-list-with-next-and-random.html