Remove Duplicates from Sorted List II

Remove Duplicates from Sorted List II

问题：

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

思路：

　　prepre pre cur 指针的相互作用

我的代码：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)   return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prepre = dummy;
        ListNode pre = head;
        ListNode cur = head.next;
        while(cur != null)
        {
            if(cur.val == pre.val)
            {
                while(cur != null && cur.val == pre.val)
                {
                    cur = cur.next;
                }
                prepre.next = cur;
                pre = cur;
                if(cur != null)
                    cur = cur.next;
                else
                    cur = null;
            }
            else
            {
                prepre = pre;
                pre = cur;
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}

学习之处：

• 这种问题无非是用两个指针标示前后位置或者三个指针标示，标示好了位置前后关系后，再画画图就差不多了。
• 当然别人也可以通过不用标志位置前后关系，也能做出来，如Remove Duplicates from Sorted List ，有人写出了这样简洁的代码，通过动还是不动的方法，简洁明了。

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return null;
        }

        ListNode node = head;
        while (node.next != null) {
            if (node.val == node.next.val) {
                node.next = node.next.next;
            } else {
                node = node.next;
            }
        }
        return head;
    }
}