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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.
1public class ListNode {
2 public int val;
3 public ListNode next;
4 public ListNode(int val=0, ListNode next=null) {
5 this.val = val;
6 this.next = next;
7 }
8}
9
10public class Solution {
11 public ListNode MergeInBetween(ListNode list1, int a, int b, ListNode list2) {
12 ListNode prevA = list1;
13 for (int i = 0; i < a - 1; i++) {
14 prevA = prevA.next;
15 }
16 ListNode nodeB = prevA;
17 for (int i = 0; i < b - a + 2; i++) {
18 nodeB = nodeB.next;
19 }
20 prevA.next = list2;
21 ListNode tail = list2;
22 while (tail.next != null) {
23 tail = tail.next;
24 }
25 tail.next = nodeB;
26 return list1;
27 }
28}This C# solution traverses list1 using two pointers, similar to the previous solutions. It finds a position before and after where list2 should be inserted, ensuring the rest of list1 remains connected.
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.
1C does not naturally support recursive list manipulations easily, especially without introducing significant complications. Therefore, this approach is better showcased in languages with better support for recursive structures.