Table of Contents
Task
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 representingNode.val
random_index
: the index of the node (range from0
ton-1
) that therandom
pointer points to, ornull
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
-10
4<= Node.val <= 10
4Node.random
isnull
or is pointing to some node in the linked list.
This problem was taken from https://leetcode.com/problems/copy-list-with-random-pointer/
Solution
Brute force using tree traversal
/** * // Definition for a Node. * function Node(val, next, random) { * this.val = val; * this.next = next; * this.random = random; * }; */ /** * @param {Node} head * @return {Node} */ var copyRandomList = function(head) { function dfsTraverse(node, visited={}) { if(!node) { return null; } // if a new node is created, return it. Otherwise you will fall into circular loops if(node?.clone) { return node?.clone; } var newNode = new Node(node?.val, null, null); node.clone = newNode; var next = dfsTraverse(node?.next); var random = dfsTraverse(node?.random); newNode.next = next; newNode.random = random; return newNode; } var result = dfsTraverse(head); return result; }; function Node(val, next, random) { this.val = val; this.next = next; this.random = random; } ;// [1,null],[2,0],[3,1] var nodes = {}; nodes['1'] = new Node(1,null,null); nodes['2'] = new Node(2,null,null); nodes['3'] = new Node(3,null,null); nodes['1'].next = nodes['2']; nodes['1'].random = null; nodes['2'].next = nodes['3']; nodes['2'].random = nodes['1']; nodes['3'].next = null; nodes['3'].random = nodes['2']; //console.log("root"); //console.log(nodes['7']); var result = copyRandomList(nodes['1']); console.log(result);
More elegant solution
/** * // Definition for a Node. * function Node(val, next, random) { * this.val = val; * this.next = next; * this.random = random; * }; */ /** * @param {Node} head * @return {Node} */ var copyRandomList = function(head) { let cur = head; const copy = new Map(); // add all new nodes and values for now while (cur) { copy.set(cur, new Node(cur.val)); cur = cur.next; } cur = head; // iterate again and point curent node to the newly created nodes using the key while (cur) { copy.get(cur).next = copy.get(cur.next) || null; copy.get(cur).random = copy.get(cur.random) || null; cur = cur.next; } return copy.get(head); } function Node(val, next, random) { this.val = val; this.next = next; this.random = random; }; // [1,null],[2,0],[3,1] var nodes = {}; nodes['1'] = new Node(1,null,null); nodes['2'] = new Node(2,null,null); nodes['3'] = new Node(3,null,null); nodes['1'].next = nodes['2']; nodes['1'].random = null; nodes['2'].next = nodes['3']; nodes['2'].random = nodes['1']; nodes['3'].next = null; nodes['3'].random = nodes['2']; var result = copyRandomList(nodes['1']); console.log(result);