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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 <iostream>
2
3struct ListNode {
4 int val;
5 ListNode *next;
6 ListNode(int x) : val(x), next(NULL) {}
7};
8
9ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {
10 ListNode* prevA = list1;
11 for (int i = 0; i < a - 1; ++i) {
12 prevA = prevA->next;
13 }
14 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 != nullptr) {
20 list2 = list2->next;
21 }
22 list2->next = nodeB;
23 return list1;
24}In this C++ solution, we traverse list1 similarly to the C solution. We locate the node immediately before a and the node immediately after b. Then, link the node at a-1 with the head of list2 and the end of list2 with the first node after b.
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.