Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
有序链表去重，两个指针，一次遍历，一个指针tail1指向原链表尾部，一个指针tail2指向去重后的链表尾部，如果tail1的val和tail2的val不一样，则说名这个没出现过，就把这个tail1的val移动到tail2的下一个节点的val上。
/**
* 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 head;
ListNode* tail1 = head->next;
ListNode* tail2 = head;
while(tail1 != NULL)
{
if(tail1->val != tail2->val)
{
tail2->next->val = tail1->val;
tail2 = tail2->next;
}
tail1 = tail1->next;
}
tail2->next = NULL;
return head;
}
};
