1669. Merge in Between Linked Lists - Explanation

# 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:
        cur = list1
        arr = []
        while cur:
            arr.append(cur)
            cur = cur.next
        arr[a - 1].next = list2

        cur = list2
        while cur.next:
            cur = cur.next

        cur.next = arr[b + 1]
        return list1

# 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:
        curr = list1
        i = 0

        while i < a - 1:
            curr = curr.next
            i += 1
        head = curr

        while i <= b:
            curr = curr.next
            i += 1

        head.next = list2

        while list2.next:
            list2 = list2.next
        list2.next = curr

        return list1

# 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:
        if a == 1 :
            nxt = list1.next
            list1.next = list2
            while list2.next:
                list2 = list2.next
            self.mergeInBetween(nxt, 0, b - 1, list2)
            return list1

        if b == 0:
            list2.next = list1.next
            return list1

        self.mergeInBetween(list1.next, a - 1, b - 1, list2)
        return list1