LeetCode之“链表”：Remove Duplicates from Sorted List && Remove Duplicates from Sorted List II

　 1. Remove Duplicates from Sorted List

　　题目链接

　　题目要求：

　　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.

　　这道题不难。具体程序如下：

 /**
  * 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 || !head->next)
             return head;

         int preVal = head->val;
         ListNode *preNode = head;
         ListNode *start = head->next;
         while(start)
         {
             while(start && start->val == preNode->val)
             {
                 ListNode *next = start->next;
                 preNode->next = next;
                 delete start;
                 start = next;
             }

             if(!start)
                 break;

             preNode = start;
             start = start->next;
         }

         return head;
     }
 };

　 2. Remove Duplicates from Sorted List II

　　题目链接

　　题目要求：

　　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.

　　该题可以通过添加dummy节点以方便编程。具体程序如下：

 /**
  * 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 || !head->next)
             return head;

         int preVal = head->val;
         ListNode *dummy = new ListNode();
         dummy->next = head;
         ListNode *prepreNode = dummy, *preNode = head;
         ListNode *start = head->next;
         while(start)
         {
             bool flag = true;
             while(start && start->val == preNode->val)
             {
                 flag = false;
                 ListNode *next = start->next;
                 preNode->next = next;
                 delete start;
                 start = next;
             }

             if(flag)
             {
                 prepreNode = preNode;
                 preNode = preNode->next;
             }
             else
             {
                 prepreNode->next = preNode->next;
                 delete preNode;
                 preNode = prepreNode->next;
             }

             if(!preNode || !preNode->next)
                 break;
             start = preNode->next;
         }

         head = dummy->next;
         delete dummy;
         dummy = nullptr;

         return head;
     }
 };