删除链表中重复的节点：

首先删除链表的节点我们必然需要至少两个指针，跑在前面的叫head，后面的叫post。首先把遍历整个链表的代码(不加删除节点)写出来：

private HashSet<Integer> set = new HashSet<>();
    public ListNode deleteDuplication(ListNode pHead) {
        ListNode head = pHead;
        ListNode post = pHead;
        while(head != null){
            set.add(head.val);
            head = head.next;
            post.next = head;
            post = post.next;
        }
        return pHead;
    }

然后遇到重复的节点只需要head = head.next，删除节点时只需要 post.next = head即可。所以代码的雏形就出来了：（注意，这里只是去掉多余的重复节点即1->2->3->4->5，而题中要求去掉所有重复节点1->2->5）。

    private HashSet<Integer> set = new HashSet<>();
    public ListNode deleteDuplication(ListNode pHead) {
        ListNode head = pHead;
        ListNode post = pHead;
        while(head != null){
            set.add(head.val);
            head = head.next;

            while(head != null && set.contains(head.val)){
                head = head.next;
            }
            post.next = head;
            post = post.next;
        }
        return pHead;
    }

所以我们需要两个set用来找到重复的节点结合。代码如下：

private HashSet<Integer> setAll = new HashSet<>();
private HashSet<Integer> set = new HashSet<>();

while(head != null){
    if(setAll.contains(head.val))
        set.add(head.val);
    setAll.add(head.val);
    head = head.next;
}

剩余的代码和前面一样，只不过由于第一个节点也有可能是重复节点，所以需要多一次过滤，从第一个不是重复的节点开始。完整的代码如下：

    private HashSet<Integer> setAll = new HashSet<>();
    private HashSet<Integer> set = new HashSet<>();
    public ListNode deleteDuplication(ListNode pHead) {
        ListNode head = pHead;
        ListNode post = pHead;

        while(head != null){
            if(setAll.contains(head.val))
                set.add(head.val);
            setAll.add(head.val);
            head = head.next;
        }

        head = pHead;
        while(head != null && set.contains(head.val)){
            head = head.next;
        }
        pHead = head;
        post = pHead;
        while(head != null){
            head = head.next;
            while(head != null && set.contains(head.val)){
                head = head.next;
            }
            post.next = head;
            post = post.next;
        }
        return pHead;
    }