Remove Duplicates from Sorted List II

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

Example 1:
Input: 1->2->3->3->4->4->5
Output: 1->2->5
Example 2:
Input: 1->1->1->2->3
Output: 2->3

方法一：先找到重复的那个数，把和他一样的后面的数全去掉

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null) return head;

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;

        while(head.next != null && head.next.next != null) {
            if(head.next.val == head.next.next.val) {
                int val = head.next.val;
                while(head.next != null && head.next.val == val) {
                    head.next = head.next.next;
                }
            }else head = head.next;
        }
        return dummy.next;
    }
}

方法二：三个指针，real指针在对的情况下才前进(数字与前与后都不同)

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode dummy = new ListNode(0);
        ListNode real, pre, cur;
        real = dummy;
        pre = real;
        cur = head;
        while(cur != null) {
            if((pre == dummy || pre.val != cur.val) && (cur.next == null || cur.val != cur.next.val)) {
                real.next = cur;
                real = cur;
            }
            pre = cur;
            cur = cur.next;
            pre.next = null;
        }
        return dummy.next;
    }
}