Question
Given a linked list, and an input n, write a function that returns the nth-to-last element of the linked list.
eg.
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list = 1 -> 2 -> 3 -> 4 -> 5 -> null nthToLast(list, 0) = 5 nthToLast(list, 1) = 4 nthToLast(list, 4) = 1 nthToLast(list, 5) = null |
Once you think that you’ve solved the problem, click below to see the solution.
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Solution
How was that problem? You can check out the solution in the video below.
Here is the source code for the solution shown in the video (Github):
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// Private node class private class Node { private int value; private Node next; private Node(int value) { this.value = value; } } public Node nthToLast(Node node, int n) { Node curr = node; Node follower = node; // Iterate curr forward by n. If you reach the end of the list then it is // shorter than n, so you can't possible have an nth-to-last node for (int i = 0; i < n; i++) { if (curr == null) return null; curr = curr.next; } // If length is exactly n, the nth-to-last node would be null if (curr == null) return null; // Move both nodes forward in unison until curr is at the end of the list while (curr.next != null) { curr = curr.next; follower = follower.next; } return follower; } |
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