Sponsored
Sponsored
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.
1function ListNode(val, next) {
2 this.val = (val===undefined ? 0 : val)
3 this.next = (next===undefined ? null : next)
4}
5
6var mergeInBetween = function(list1, a, b, list2) {
7 let prevA = list1;
8 for (let i = 0; i < a - 1; i++) {
9 prevA = prevA.next;
10 }
11 let nodeB = prevA;
12 for (let i = 0; i < b - a + 2; i++) {
13 nodeB = nodeB.next;
14 }
15 prevA.next = list2;
16 let tail = list2;
17 while (tail.next !== null) {
18 tail = tail.next;
19 }
20 tail.next = nodeB;
21 return list1;
22};This JavaScript solution uses a similar pointer-based approach as other solutions, traversing list1 until the points where list2 connects before and after the replace segment.
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 Python solution uses recursion and reduces the problem size with each call, moving closer to the base case where list2 is directly connected.