Write a function named transferEvens that extracts even-indexed elements out of a linked list.
Your function should accept a references to a ListNode pointer to the front of a linked list.
You should remove the even-indexed elements from the first linked list and put them into a new second linked list, in the same relative order, which is then returned.
For example, suppose we have a pointer named list1 that points to the front of the following list of elements:
index 0 1 2 3 4 5 6 7 8 9 10
list1: {3, 1, 4, 15, 9, 2, 6, 5, 35, 89, 66}
If the list were set up in the following way and then your function were called:
ListNode* list2 = transferEvens(list1);
Then the final state of the two lists would be the following:
index 0 1 2 3 4 5
list1: {1, 15, 2, 5, 89}
list2: {3, 4, 9, 6, 35, 66}
Constraints:
Do not construct any new ListNode objects in solving this problem (though you may create as many ListNode* pointer variables as you like).
Do not use any auxiliary data structures to solve this problem (no array, vector, stack, queue, string, etc).
Assume that you are using the ListNode structure as defined below:
struct ListNode {
int data; // value stored in each node
ListNode* next; // pointer to next node in list (nullptr if none)
}
