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:新建一个链表，遍历原来链表，若不重复则将结点插入新链表，重复则跳过.为了避免遍历最后为重复结点，最后需将新链表末尾置NULL

#include<iostream>#include<vector>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)  //如果不重复，插入{TmpResultNode->next = NorepeatNode;TmpResultNode = TmpResultNode->next;}NorepeatNode  = NorepeatNode  -> next;}TmpResultNode->next = NULL;       //尾部置NULLreturn ResultList->next;}