Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 2 | Linked Lists
2.1 Write code to remove duplicates from an unsorted linked list.
FOLLOW UP
How would you solve this problem if a temporary buffer is not allowed?
pg 50
SOLUTION
If we can use a buffer, we can keep track of elements in a hashtable and remove any dups:
1 public static void deleteDups(LinkedListNode n) {
2 Hashtable table = new Hashtable();
3 LinkedListNode previous = null;
4 while (n != null) {
5 if (table.containsKey(n.data)) previous.next = n.next;
6 else {
7 table.put(n.data, true);
8 previous = n;
9 }
10 n = n.next;
11 }
12 }
Without a buffer, we can iterate with two pointers: “current” does a normal iteration, while
“runner” iterates through all prior nodes to check for dups. Runner will only see one dup
per node, because if there were multiple duplicates they would have been removed already.
1 public static void deleteDups2(LinkedListNode head) {
2 if (head == null) return;
3 LinkedListNode previous = head;
4 LinkedListNode current = previous.next;
5 while (current != null) {
6 LinkedListNode runner = head;
7 while (runner != current) { // Check for earlier dups
8 if (runner.data == current.data) {
9 LinkedListNode tmp = current.next; // remove current
10 previous.next = tmp;
11 current = tmp; // update current to next node
12 break; // all other dups have already been removed
13 }
14 runner = runner.next;
15 }
16 if (runner == current) { // current not updated - update now
17 previous = current;
18 current = current.next;
19 }
20 }
21 }
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Cracking the Coding Interview | Data Structures