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This approach involves using two pointers to identify the nodes in list1 where list2 will be inserted.
First, traverse list1 to reach the node right before the ath node. Next, continue to traverse to just beyond the bth node. The node at this position serves as the connection to the remainder of list1 after the insertion point.
Finally, link the identified nodes with list2.
Time Complexity: O(n + m) where n is the number of nodes in list1 up to b and m is the length of list2. Space Complexity: O(1) as we use a constant amount of auxiliary space.
1class ListNode:
2 def __init__(self, val=0, next=None):
3 self.val = val
4 self.next = next
5
6def mergeInBetween(list1, a, b, list2):
7 prevA = list1
8 for _ in range(a - 1):
9 prevA = prevA.next
10 nodeB = prevA
11 for _ in range(b - a + 2):
12 nodeB = nodeB.next
13 prevA.next = list2
14 tail = list2
15 while tail.next:
16 tail = tail.next
17 tail.next = nodeB
18 return list1This Python solution performs similar operations to other solutions. Pointers prevA and nodeB help link the non-replaced parts of list1 with list2.
This approach solves the problem recursively, leveraging recursive traversal to reach the split points on list1 and link list2 in the required section of list1.
Base cases are established for when the algorithm should stop the recursive calls and perform actions to connect the segments together.
Complexity details are not provided for C in recursive solutions.
1
This JavaScript solution uses recursion to simplify the connections between the nodes. Specifically, it recurses down to the point of insertion and builds back up from there.