Cracking the Coding Interview Q2.1

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?

思路1：用hashset存储已经出现的节点，如果重复则删除。空间O(N)，时间O(N)。

思路2：两个指针，对于每个当前节点，遍历后面的元素删除重复。 空间O(1)，时间O(n^2)。

思路3：移动窗口法是否可行？

public static void deleteDupsA(LinkedListNode n) {
HashSet<Integer> set = new HashSet<Integer>();
LinkedListNode previous = null;
while (n != null) {
if (set.contains(n.data)) {
previous.next = n.next;
} else {
set.add(n.data);
previous = n;
}
n = n.next;
}
}

public static void deleteDupsC(LinkedListNode head) {
if (head == null) return;
LinkedListNode previous = head;
LinkedListNode current = previous.next;
while (current != null) {
// Look backwards for dups, and remove any that you see.
LinkedListNode runner = head;
while (runner != current) {
if (runner.data == current.data) {
LinkedListNode tmp = current.next;
previous.next = tmp;
current = tmp;
/* We know we can't have more than one dup preceding
* our element since it would have been removed
* earlier. */
break;
}
runner = runner.next;
}

/* If runner == current, then we didn't find any duplicate
* elements in the previous for loop.  We then need to
* increment current.
* If runner != current, then we must have hit the �break�
* condition, in which case we found a dup and current has
* already been incremented.*/
if (runner == current) {
previous = current;
current = current.next;
}
}
}

public static void deleteDupsB(LinkedListNode head) {
if (head == null) return;

LinkedListNode current = head;
while (current != null) {
/* Remove all future nodes that have the same value */
LinkedListNode runner = current;
while (runner.next != null) {
if (runner.next.data == current.data) {
runner.next = runner.next.next;
} else {
runner = runner.next;
}
}
current = current.next;
}
}