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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        head = dummy;
        while (head.next != null && head.next.next != null) {
            if (head.next.val == head.next.next.val) {
                // Should keep the value of the head of dup nodes.
                // head.next.val == value is the condition which triggers pointer sliding.
                int value = head.next.val;
                while (head.next != null && head.next.val == value) {
                    head.next = head.next.next;
                }
            } else { // Should use else here. Because when head.next = head.next.next, it will move forward itself.
                head = head.next;
            }
        }
        return dummy.next;
    }
}
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