Merge-Two-Linked-List.md

status	tags	Link	sr-due	sr-interval	sr-ease
	LC_Easy Linked-List NeetCode150		2024-09-30	1	230

Couldn't solve it in 10 minutes, watching coding portion and try again.

Okay, finish the actual code after watching for 3 minutes. Only thing missing was curr = curr.next oops.

Problem name

title: Psuedo Solution
collapse: closed
- First point
- Second point

Problem statement

Solution (Optimized)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        newList = ListNode()

        curr = newList

        currList1 = list1
        currList2 = list2
        while currList1 and currList2:
            if currList1.val < currList2.val:
                curr.next = currList1
                currList1 = currList1.next
            else:
                curr.next = currList2
                currList2 = currList2.next
            curr = curr.next

        if currList1:
            curr.next = currList1

        if currList2:
            curr.next = currList2

        return newList.next

You don't have to make currList1 or 2, if list1 and list2 can be changed.

---

Cue Flashcards 🗃

---