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.

解題思路:新建一個鏈表，遍歷原來鏈表，若不重復則將結點插入新鏈表，重復則跳過.為了避免遍歷最後為重復結點，最後需將新鏈表末尾置NULL

#include
#include
using namespace std;
//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
ListNode *deleteDuplicates(ListNode *head) {
	ListNode*ResultList = new ListNode(0);
	ListNode*TmpResultNode = ResultList;
	ListNode*NorepeatNode = head;
	while (NorepeatNode != NULL){
		ListNode*PreNode = NorepeatNode;
		while (NorepeatNode->next != NULL&&NorepeatNode->val == NorepeatNode->next->val)
			NorepeatNode = NorepeatNode->next;
		if (NorepeatNode == PreNode)  //如果不重復，插入
		{
			TmpResultNode->next = NorepeatNode;
			TmpResultNode = TmpResultNode->next;
		}
		NorepeatNode  = NorepeatNode  -> next;
	}
	TmpResultNode->next = NULL;       //尾部置NULL
	return ResultList->next;
}