A linked list of length n
is given such that each node contains an additional random pointer, which could point to any node in the list, or null
.
Construct a deep copy of the list. The deep copy should consist of exactly n
brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next
and random
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. None of the pointers in the new list should point to nodes in the original list.
For example, if there are two nodes X
and Y
in the original list, where X.random --> Y
, then for the corresponding two nodes x
and y
in the copied list, x.random --> y
.
Return the head of the copied linked list.
The linked list is represented in the input/output as a list of n
nodes. Each node is represented as a pair of [val, random_index]
where:
val
: an integer representing Node.val
random_index
: the index of the node (range from 0
to n-1
) that the random
pointer points to, or null
if it does not point to any node.
Your code will only be given the head
of the original linked list.
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
Node.random
is null
or is pointing to some node in the linked list.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.
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;
}
}
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.