Code-Memo

Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists.

Approach

Iterative Approach

1. Initialization:
   • Initialize a dummy node to simplify the code by avoiding edge cases when appending nodes.
   • Use current to point to the current node in the merged list, starting from dummy.
2. Merge Process:
   • While both list1 and list2 are not None:
     • Compare the values of the current nodes of list1 and list2.
     • Append the smaller node to current.next and move current to current.next.
     • Move list1 or list2 forward depending on which node was appended.
3. Handle Remaining Nodes:
   • After exiting the loop, append the remaining nodes of list1 or list2 to the end of the merged list.
4. Return Result:
   • Return dummy.next, which points to the head of the merged sorted list.

Time Complexity

• The time complexity of this approach is ( O(m + n) ), where ( m ) and ( n ) are the lengths of list1 and list2, respectively. We iterate through each list once.

class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(0)
        current = dummy

        while list1 and list2:
            if list1.val <= list2.val:
                current.next = list1
                list1 = list1.next
            else:
                current.next = list2
                list2 = list2.next
            current = current.next

        if list1:
            current.next = list1
        elif list2:
            current.next = list2

        return dummy.next