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3. Merge Two Sorted Lists

easyAsked at Box

Merge two sorted linked lists into one sorted list — Box uses this when merging two sorted streams of file change events during sync reconciliation.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

You are given the heads of two sorted linked lists list1 and list2. Merge them into a single sorted list by splicing together the nodes of the first two lists, and return the head of the merged list.

Constraints

  • 0 <= list size <= 50
  • -100 <= Node.val <= 100
  • Both lists are sorted non-decreasing

Examples

Example 1

Input
list1 = [1,2,4], list2 = [1,3,4]
Output
[1,1,2,3,4,4]

Example 2

Input
list1 = [], list2 = [0]
Output
[0]

Approaches

1. Brute force collect-and-sort

Push all values into an array, sort, rebuild list.

Time
O(n log n)
Space
O(n)
const arr = [];
while (list1) { arr.push(list1.val); list1 = list1.next; }
while (list2) { arr.push(list2.val); list2 = list2.next; }
arr.sort((a,b) => a-b);
// rebuild list from arr

Tradeoff:

2. Two-pointer splice

Use a dummy head; at each step pick the smaller current node and advance.

Time
O(n+m)
Space
O(1)
function mergeTwoLists(l1, l2) {
  const dummy = { next: null };
  let tail = dummy;
  while (l1 && l2) {
    if (l1.val <= l2.val) { tail.next = l1; l1 = l1.next; }
    else { tail.next = l2; l2 = l2.next; }
    tail = tail.next;
  }
  tail.next = l1 || l2;
  return dummy.next;
}

Tradeoff:

Box-specific tips

Box prizes the dummy-head pattern — it shows you understand in-place pointer manipulation, which they map to splicing file-change journals without re-allocating buffers.

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