Gyh's Braindump

138. Copy List with Random Pointer

tags
LinkedList, Recursion, trick
source
leetcode

Edge Cases

nodes can have same value

Solution 1 - Recursion

class Solution {
    public Node copyRandomList(Node head) {
        if (head == null) return null;

        Map<Node, Node> clonedNodes = new HashMap<>();
        return copyRandomListHelper(head, clonedNodes);
    }

    private Node copyRandomListHelper(Node head, Map<Node, Node> clonedNodes) {
        if (head == null) return null;
        if (clonedNodes.containsKey(head)) return clonedNodes.get(head);

        Node newNode = new Node(head.val);
        clonedNodes.put(head, newNode);
        newNode.next = copyRandomListHelper(head.next, clonedNodes);
        newNode.random = copyRandomListHelper(head.random, clonedNodes);

        return newNode;
    }
}

Complexity

  • time: O(N)
  • space: O(N)

Solution 2 - HashMap, one pass

public class Solution {
  // Visited dictionary to hold old node reference as "key" and new node reference as the "value"
  HashMap<Node, Node> visited = new HashMap<Node, Node>();

  public Node getClonedNode(Node node) {
    // If the node exists then
    if (node != null) {
      // Check if the node is in the visited dictionary
      if (this.visited.containsKey(node)) {
        // If its in the visited dictionary then return the new node reference from the dictionary
        return this.visited.get(node);
      } else {
        // Otherwise create a new node, add to the dictionary and return it
        this.visited.put(node, new Node(node.val, null, null));
        return this.visited.get(node);
      }
    }
    return null;
  }

  public Node copyRandomList(Node head) {

    if (head == null) {
      return null;
    }

    Node oldNode = head;

    // Creating the new head node.
    Node newNode = new Node(oldNode.val);
    this.visited.put(oldNode, newNode);

    // Iterate on the linked list until all nodes are cloned.
    while (oldNode != null) {
      // Get the clones of the nodes referenced by random and next pointers.
      newNode.random = this.getClonedNode(oldNode.random);
      newNode.next = this.getClonedNode(oldNode.next);

      // Move one step ahead in the linked list.
      oldNode = oldNode.next;
      newNode = newNode.next;
    }
    return this.visited.get(head);
  }
}

Complexity

  • time: O(N)
  • space: O(N)

Solution 3 - O(1) space

  1. First pass: create cloned nodes and append to original node
  1. assign random of cloned nodes
  2. assemble cloned nodes

Complexity

  • time: O(N)
  • space: O(1)