题解 | #删除链表中重复的结点#

/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {
    public ListNode deleteDuplication(ListNode pHead) {
        ListNode head = new ListNode(-1);
        head.next = pHead;
        // 上⼀个元素指针为pre
        ListNode pre = head, cur = pHead;
        while (cur != null) {
            if (cur.next != null && cur.val == cur.next.val) {
                cur = cur.next;
                while (cur.next != null && cur.val == cur.next.val) {
                    cur = cur.next;
                }
                cur = cur.next;
                pre.next = cur;
            } else {
                pre = cur;
                cur = cur.next;
            }
        }
        return head.next;
    }
}

解题思想：

* 方式一：借助map统计次数，然后遍历形成新链表（排除重复链表节点）

* 方式二：自身遍历去除重复节点，改变指针指向