Task Description A linked list of length is given such that each node contains an additional random pointer, which could point to any node in the list, or . n null Construct a of the list. The deep copy should consist of exactly nodes, where each new node has its value set to the value of its corresponding original node. Both the and pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. . deep copy n brand new next random None of the pointers in the new list should point to nodes in the original list For example, if there are two nodes and in the original list, where , then for the corresponding two nodes and in the copied list, . X Y X.random --> Y x y x.random --> y Return . the head of the copied linked list The linked list is represented in the input/output as a list of nodes. Each node is represented as a pair of where: n [val, random_index] : an integer representing val Node.val : the index of the node (range from to ) that the pointer points to, or if it does not point to any node. random_index 0 n-1 random null Your code will be given the of the original linked list. only head Example 1: Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]] Example 2: Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]] Example 3: Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]] Constraints: 0 <= n <= 1000 -104 <= Node.val <= 104 is or is pointing to some node in the linked list. Node.random null Reasoning: For the first glance this task looks like hard one although in reality it isn’t. But it has one property which we need to pay attention to and build a habit around it. Whenever you see a problem which can be divided into smaller problems — divide it. It’s one of the essential patterns you can apply to almost any problem. Let me help you with this. Let’s reread the task again and divide it into smaller subproblems. We need to create a copy of each node with the same value as original one. We need to set pointers to the next node the same way as in original one We need to set pointers to the random node the same way as in original one+ You can say at this point, wait instead of having one problem now we have 3. Yes you’re right but they are much easier to solve. Let’s jump to the solution section and I’ll prove it to you. Prerequisite: If you want to debug your code, which I personally recommend, you must introduce this class class Node
{
   int val;
   Node next;
   Node random;

   public Node(int val)
   {
      this.val = val;
      this.next = null;
      this.random = null;
   }
} Solution: Let’s start from solving the first subproblem. Let’s do simple thing iterate through linked list and create a copy for every node in it. I choose HashMap for storing pairs like old node -> new node. As long as we’re going to iterate from head of the list to tail let’s also introduce temp variable and set it to be the head of the list. Map<Node, Node> map = new HashMap<>();
Node temp = head;
while (temp != null)
{
   map.put(temp, new Node(temp.val));
   temp = temp.next;
} We solved the first subproblem. You’ll be surprised how easy we can solve the next two. We already have connections between old nodes and new ones. We only need to populate pointers to next node and random pointer. Our map already stores all the information we need. Let’s again set temp to be the head of the list. Let’s also introduce new dummy node which will be storing link to our freshly copied head. The last thing we need is prev pointer which will help us to set up relationship node -> next node Our solutions should look like this public Node copyRandomList(Node head)
{
   Map<Node, Node> map = new HashMap<>();
   Node temp = head;
   while (temp != null)
   {
      map.put(temp, new Node(temp.val));
      temp = temp.next;
   }

   temp = head;
   Node dummy = new Node(0);
   Node prev = dummy;
   while (temp != null)
   {
      prev.next = map.get(temp);
      map.get(temp).random = map.get(temp.random);

      prev = prev.next;
      temp = temp.next;
   }

   return dummy.next;
} The code above gives us linear time and space complexity. Also published . here

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Java Algorithms: Copying List with Random Pointer (LeetCode)

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