用于在 O(1) 空间中克隆具有 Next 和随机指针的链表的Java程序

给定一个链表，每个节点都有两个指针。第一个指向列表的下一个节点，但是，另一个指针是随机的，可以指向列表的任何节点。编写一个程序，在 O(1) 空间中克隆给定列表，即没有任何额外空间。
例子：

Input : Head of the below-linked list

Output :
A new linked list identical to the original list.

在之前的帖子中，Set-1 和 Set-2 讨论了各种方法，并且还提供了 O(n) 空间复杂度实现。
在这篇文章中，我们将实现一个不需要额外空间的算法，如 Set-1 中所讨论的。
下面是算法：

创建节点 1 的副本并将其插入到原始链表中的节点 1 和节点 2 之间，创建 2 的副本并将其插入到 2 和 3 之间。以这种方式继续，在第 N 个节点之后添加 N 的副本
现在以这种方式复制随机链接

original->next->random= original->random->next;  /*TRAVERSE 
TWO NODES*/

这是有效的，因为 original->next 只不过是原始的副本，而 Original->random->next 只不过是随机的副本。
现在以这种方式在一个循环中恢复原始和复制链表。

original->next = original->next->next;
     copy->next = copy->next->next;

确保 original->next 为 NULL 并返回克隆列表

下面是实现。

Java

// Java program to clone a linked list with next
// and arbit pointers in O(n) time
class GfG {
  
    // Structure of linked list Node
    static class Node {
        int data;
        Node next, random;
        Node(int x)
        {
            data = x;
            next = random = null;
        }
    }
  
    // Utility function to print the list.
    static void print(Node start)
    {
        Node ptr = start;
        while (ptr != null) {
            System.out.println("Data = " + ptr.data
                               + ", Random = "
                               + ptr.random.data);
            ptr = ptr.next;
        }
    }
  
    // This function clones a given
    // linked list in O(1) space
    static Node clone(Node start)
    {
        Node curr = start, temp = null;
  
        // insert additional node after
        // every node of original list
        while (curr != null) {
            temp = curr.next;
  
            // Inserting node
            curr.next = new Node(curr.data);
            curr.next.next = temp;
            curr = temp;
        }
        curr = start;
  
        // adjust the random pointers of the
        // newly added nodes
        while (curr != null) {
            if (curr.next != null)
                curr.next.random = (curr.random != null)
                                       ? curr.random.next
                                       : curr.random;
  
            // move to the next newly added node by
            // skipping an original node
            curr = curr.next.next;                    
        }
  
        Node original = start, copy = start.next;
  
        // save the start of copied linked list
        temp = copy;
  
        // now separate the original list and copied list
        while (original != null) {
            original.next =original.next.next;
  
          copy.next = (copy.next != null) ? copy.next.next
                                            : copy.next;
            original = original.next;
            copy = copy.next;
        }
        return temp;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        Node start = new Node(1);
        start.next = new Node(2);
        start.next.next = new Node(3);
        start.next.next.next = new Node(4);
        start.next.next.next.next = new Node(5);
  
        // 1's random points to 3
        start.random = start.next.next;
  
        // 2's random points to 1
        start.next.random = start;
  
        // 3's and 4's random points to 5
        start.next.next.random = start.next.next.next.next;
        start.next.next.next.random
            = start.next.next.next.next;
  
        // 5's random points to 2
        start.next.next.next.next.random = start.next;
  
        System.out.println("Original list : ");
        print(start);
  
        System.out.println("Cloned list : ");
        Node cloned_list = clone(start);
        print(cloned_list);
    }
}
  
// This code is contributed by Prerna Saini.

输出

Original list : 
Data = 1, Random  = 3
Data = 2, Random  = 1
Data = 3, Random  = 5
Data = 4, Random  = 5
Data = 5, Random  = 2

Cloned list : 
Data = 1, Random  = 3
Data = 2, Random  = 1
Data = 3, Random  = 5
Data = 4, Random  = 5
Data = 5, Random  = 2

有关更多详细信息，请参阅有关在 O(1) 空间中使用 next 和随机指针克隆链接列表的完整文章！