n elements in singly linked list
- by Codenotguru
The following function is trying to find the nth to last element of a singly linked list.
for ex: if the elements are 8-10-5-7-2-1-5-4-10-10 then the result is
7th to last node is 7.
Can anybody help me on how this code is working or is there a better and simpler approach?
LinkedListNode nthToLast(LinkedListNode head, int n) {
if (head == null || n < 1) {
return null;
}
LinkedListNode p1 = head;
LinkedListNode p2 = head;
for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead
if (p2 == null) {
return null; // not found since list size < n
}
p2 = p2.next;
}
while (p2.next != null) {
p1 = p1.next;
p2 = p2.next;
}
return p1;
}
