1669 Merge in Between Linked Lists

Approach 1

Nothing fancy, just some traversal and pointer manipulation.

• Initialize two dummy header nodes
• Initialize curr and prev pointers for both list1 and list2
• Advancing both prev and curr in list1. Traverse such that prev is at the node before a and curr is at the node a
• Advancing only curr in list1 it's at the next of node of b. If b == 4, curr is at the node 5
• Advancing prev and curr in list2 until curr is at the end of list2
• connection time!!

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
        dummy1,dummy2 = ListNode(0,list1),ListNode(0,list2)
        curr1,curr2 = list1,list2
        prev1,prev2 = dummy1,dummy2

        counter = 0
        while counter < a and curr1:
            curr1 = curr1.next
            prev1 = prev1.next
            counter += 1

        # when we got here
        while counter <= b:
            curr1 = curr1.next
            counter += 1

        while curr2:
            prev2 = prev2.next
            curr2 = curr2.next

        # connect
        prev1.next = list2
        prev2.next = curr1

        return dummy1.next