LeetCode 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.

For example,
Given

1->2->3->3->4->4->5

, return

1->2->5

.
Given

1->1->1->2->3

, return

2->3

.

解题思路：防止情况的多样性，先在head前面加上一个节点node，这样无论什么情形都可以统一为一种处理方式。

　　设置一个work指针指向node，如果work->next->val == work->next->next->val，表明有相同元素出现，则在内循环内将相同元素删除到只剩2个，在循环结束时，再讲这2个相同的也删除，如果work->next->val ！= work->next->next->val；则跳过

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if(head==NULL)return NULL;

ListNode *node = new ListNode(0);
node->next = head;

ListNode *work = node;
int key = node->next->val;

while(work->next && work->next->next)
{
if(work->next->val == work->next->next->val)
{
ListNode *p = work->next;
while(p->next->next && p->next->val==p->next->next->val)
{
p->next = p->next->next;
}
work->next = work->next->next->next;
}
else
{
work = work->next;
}
}
return node->next;
}
};