复制含有随机指针节点的链表

复制含有随机指针节点的链表
【题目】 一种特殊的链表节点类描述如下：

public class Node { 
    
	public int value; 
    public Node next; 
    publicNode rand;
    
  public Node(int value) { 
    this.value = value; 
  }
}

Node类中的value是节点值，next指针和正常单链表中next指针的意义一 样，都指向下一个节点，rand指针是Node类中新增的指针，这个指针可 能指向链表中的任意一个节点，也可能指向null。 给定一个由Node节点类型组成的无环单链表的头节点head，请实现一个函数完成这个链表中所有结构的复制，并返回复制的新链表的头节点。

进阶：不使用额外的数据结构，只用有限几个变量，且在时间复杂度为 O(N)内完成原问题要实现的函数。

public class CopyListWithRandom {
	public static class Node {
		public int value;
		public Node next;
		public Node rand;

		public Node(int value) {
			this.value = value;
		}
	}

	public static Node copyListWithRand1(Node head) {
		HashMap<Node, Node> map = new HashMap<>();
		Node cur = head;
		while (cur != null) {
			map.put(cur, new Node(cur.value));
			cur = cur.next;
		}
		cur = head;
		while (cur != null) {
			map.get(cur).next = map.get(cur.next);
			map.get(cur).rand = map.get(cur.rand);
			cur = cur.next;
		}
		return map.get(head);
	}

	public static Node copyListWithRand2(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		Node next = null;
		// copy node and link to every node
		while (cur != null) {
			next = cur.next;
			cur.next = new Node(cur.value);
			cur.next.next = next;
			cur = next;
		}
		cur = head;
		Node curCopy = null;
		// set copy node rand
		while (cur != null) {
			next = cur.next.next;
			curCopy = cur.next;
			curCopy.rand = cur.rand != null ? cur.rand.next : null;
			cur = next;
		}
		Node res = head.next;
		cur = head;
		// split
		while (cur != null) {
			next = cur.next.next;
			curCopy = cur.next;
			cur.next = next;
			curCopy.next = next != null ? next.next : null;
			cur = next;
		}
		return res;
	}