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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4struct ListNode {
5 int val;
6 struct ListNode *next;
7};
8
9struct ListNode* mergeInBetween(struct ListNode* list1, int a, int b, struct ListNode* list2) {
10 struct ListNode* prevA = list1;
11 for (int i = 0; i < a - 1; i++) {
12 prevA = prevA->next;
13 }
14 struct ListNode* nodeB = prevA;
15 for (int i = 0; i < b - a + 2; i++) {
16 nodeB = nodeB->next;
17 }
18 prevA->next = list2;
19 while (list2->next != NULL) {
20 list2 = list2->next;
21 }
22 list2->next = nodeB;
23 return list1;
24}In this C solution, we define a struct ListNode and operate on pointers within the linked list. First, we traverse to just before the ath position in list1 (prevA), and then traverse from there to just beyond the bth node (nodeB) in list1. After that, we link prevA to list2 and traverse to the end of list2 to link it to nodeB.
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.