给定一个排好序的链表，删除链表中重复的结点，返回链表头指针。

主要解决方法代码如下：

public class DeleteRepeatNode { static ArrayList<Integer> orderList;

static public List<Integer> deleteRepeatNode(ArrayList<Integer> arrayList) throws Exception {

    if (arrayList == null) {
        throw new Exception("arrayList cant empty");
    }
    int lastNodeValue = arrayList.get(0);
    for (int position = 1; position < arrayList.size(); position++) {
        //因为是排序链表所以相等的值只有可能出现在相邻的位置，然后如果和前一个相等则依次删除
        while (lastNodeValue == arrayList.get(position)) {
            arrayList.remove(position);
        }
        lastNodeValue = arrayList.get(position);
    }
    return arrayList;
}

public static void main(String [] args)
{
    orderList=new ArrayList<Integer>();
    orderList.add(1);
    orderList.add(2);
    orderList.add(3);
    orderList.add(3);
    orderList.add(3);
    orderList.add(4);
    orderList.add(4);
    orderList.add(5);
    try {
        deleteRepeatNode(orderList);
    } catch (Exception e) {
        e.printStackTrace();
    }


}

}

# -*- coding: utf-8 -*-

'''
测试用例：
1.功能测试：重复的节点位于链表的头部/中间/尾部;链表中没有重复的节点;
2.特殊输入测试：指向链表头节点的为nullptr指针;链表中所有节点都是重复的;
'''

class ListNode:
    def __init__(self,x):
        self.val=x
        self.next=None
class Solution:
    def DeleteDuplication(self,pHead):
        if pHead==None:
            return
        preNode=None
        pNode=pHead
        while pNode!=None:
            needDelete=False
            pNext=pNode.next
            if pNext!=None and pNext.val==pNode.val:
                needDelete=True
            if needDelete==False:
                preNode=pNode
                pNode=pNode.next
            else:
                nodeVal=pNode.val
                pToBeDel=pNode
                while pToBeDel!=None and pToBeDel.val==nodeVal:
                    pToBeDel=pToBeDel.next   #这里表示对重复节点进行删除
                if preNode==None:
                    pHead=pToBeDel           #此时pToBeDel表示删除重复节点后的当前节点
                    pNode=pToBeDel
                    continue
                else:
                    preNode.next=pToBeDel
                pNode=preNode
        return pHead

#测试用例：
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(3)
node5 = ListNode(4)
node6 = ListNode(4)
node7 = ListNode(5)
node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5
node5.next = node6
node6.next = node7

s = Solution()
print(s.DeleteDuplication(node1).next.next.val)