Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

思路

用HashSet来解：对于每一个节点，如果在HashSet中存在，则删掉，若不存在，则加入HashSet。

Follow up

How would you solve this problem if a temporary buffer is not allowed?
用双指针分别指向pre和cur，双重循环，外循环循环整个链表，内循环每次从pre开始循环到链表尾部，遇到重复就删除。但是时间复杂度是O(n2)，用空间换时间

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        //用hashset来存已经存在的node，如果有重复，则remove
        Set<Integer> set = new HashSet<Integer>();
        //用前后两个node来比较，不要忘了第一个node需要加到hashset中
        ListNode cur = head.next;
        ListNode pre = head;
        set.add(pre.val);
        
        while (cur != null) {
            if (!set.contains(cur.val)) {
                set.add(cur.val);
                cur = cur.next;
                pre = pre.next;
            } else {
                cur = cur.next;
                pre.next = cur;
            }
        }
        return head;
    }
}