Leetcode 82. 删除排序链表中的重复元素 II

 1 class Solution {
 2 public:
 3     ListNode* searchPreDuplicates(ListNode *head, ListNode *p){
 4         ListNode* q = head;
 5         while(q != NULL&& q->next != NULL){
 6             if(q->next == p)
 7                 break;
 8             q = q->next;
 9         }
10         return q;
11     }
12     ListNode* deleteDuplicates(ListNode* head) {
13         ListNode* preHead= new ListNode(-1);
14         ListNode *p = preHead;
15         p->next = head;
16         p = p->next;
17         //ListNode* p = head;
18         while(p != NULL && p->next != NULL){
19             if(p->val != p->next->val){
20                 p = p->next;
21             }
22             else{
23                 ListNode* pre = searchPreDuplicates(preHead,p);
24                 
25                 while(p != NULL && p->next != NULL && p->val == p->next->val)  p = p->next;  
26                 //找重复数字的下一个节点
27                 
28                 ListNode* post = p->next;
29                 
30 
31                 pre->next = post;
32                 p = post;
33 
34             }
35         }
36         return preHead->next;
37     }
38 };

 1 class Solution {
 2 public:
 3  
 4     ListNode* deleteDuplicates(ListNode* head) {
 5         //创建哑节点
 6         ListNode *preHead = new ListNode(-1);
 7         preHead->next = head;
 8         //创建快慢指针 
 9         ListNode *slow = preHead;  //慢指针用于指向重复数字的前一个数字
10         ListNode *fast = head;   //快指针用于寻找重复数字
11 
12         while(fast != NULL && fast->next != NULL){
13             if(fast->val != fast->next->val){   //分情况讨论，如1->2和1->1->2
14                 if(slow->next == fast ) //此时之前没有重复数字
15                     slow = fast;
16                 else{
17                     slow->next = fast->next;   //此时将略过重复数字
18                 }
19             } 
20             fast = fast->next;
21         }
22         
23         if(slow->next != fast) slow->next = fast->next;
24         
25         return preHead->next;
26         
27     }
28 };