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;
        }
    }

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