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

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#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) //如果不重复，插入 { TmpResultNode->next = NorepeatNode; TmpResultNode = TmpResultNode->next; } NorepeatNode = NorepeatNode -> next; } TmpResultNode->next = NULL; //尾部置NULL return ResultList->next; }
   
  

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