readme.md

Merge Two Sorted Lists ✅

• 📁 Difficulty
  🟢 Easy

Description

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

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

Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []

Output: []

Example 3:

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

Output: [0]

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

Topics

• Linked List
• Recursion

Solution

• Time Complexity::
  • O(n+m)
• Space Complexity::
  • O(1)

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} list1
 * @param {ListNode} list2
 * @return {ListNode}
 */
var mergeTwoLists = function (list1, list2) {
  if (list1 === null) return list2
  if (list2 === null) return list1

  const merge = new ListNode(undefined);
  let pointer1 = list1, pointer2 = list2, pointerMerge = merge

  while (pointer1 !== null || pointer2 !== null) {
    if (pointer1 === null) {
      pointerMerge.val = pointer2.val
      pointer2 = pointer2?.next
    } else if (pointer2 === null) {
      pointerMerge.val = pointer1.val
      pointer1 = pointer1?.next
    } else if (pointer1.val <= pointer2.val) {
      pointerMerge.val = pointer1.val
      pointer1 = pointer1?.next
    } else {
      pointerMerge.val = pointer2.val
      pointer2 = pointer2?.next
    }

    pointerMerge.next = pointer1 !== null || pointer2 !== null ? new ListNode() : null
    pointerMerge = pointerMerge.next
  }

  return merge;
};