Cracking the Coding Interview - Fooo

Solutions to Chapter 2 | Linked Lists
2 2 Implement an algorithm to find the nth to last element of a singly linked list
SOLUTION
Note: This problem screams recursion, but we’ll do it a different way because it’s
trickier. In a question like this, expect follow up questions about the advantages
of recursion vs iteration.
Assumption: The minimum number of nodes in list is n
Algorithm:
1 Create two pointers, p1 and p2, that point to the beginning of the node
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pg 50
2 Increment p2 by n-1 positions, to make it point to the nth node from the beginning (to
make the distance of n between p1 and p2)
3 Check for p2->next == null if yes return value of p1, otherwise increment p1 and p2
If next of p2 is null it means p1 points to the nth node from the last as the distance
between the two is n
4 Repeat Step 3
1 LinkedListNode nthToLast(LinkedListNode head, int n) {
2 if (head == null || n < 1) {
3 return null;
4 }
5 LinkedListNode p1 = head;
6 LinkedListNode p2 = head;
7 for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead
8 if (p2 == null) {
9 return null; // not found since list size < n
10 }
11 p2 = p2.next;
12 }
13 while (p2.next != null) {
14 p1 = p1.next;
15 p2 = p2.next;
16 }
17 return p1;
18 }
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