LeetCode 82. Remove Duplicates from Sorted List II

原题链接在这里：https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

题目：

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.

题解：

类似Remove Duplicates from Sorted List, 不同就是要完全去掉duplicate.

当cur.next.val == cur.next.next.val 时，就一直移动dup = cur.next.next直到cur.next.val != dup.val. 然后cur的next指向dup.

Time Complexity: O(n), n 是list长度.

Space: O(1).

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode deleteDuplicates(ListNode head) {
11         if(head == null || head.next == null){
12             return head;
13         }
14         
15         ListNode dummy = new ListNode(0);
16         dummy.next = head;
17         ListNode cur = dummy;
18         while(cur.next != null && cur.next.next != null){
19             if(cur.next.val != cur.next.next.val){
20                 cur = cur.next;
21             }else{
22                 ListNode dup = cur.next.next;
23                 while(dup != null && cur.next.val == dup.val){
24                     dup = dup.next;
25                 }
26                 cur.next = dup;
27             }
28         }
29         
30         return dummy.next;
31     }
32 }

AC C++:

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode() : val(0), next(nullptr) {}
 7  *     ListNode(int x) : val(x), next(nullptr) {}
 8  *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 9  * };
10  */
11 class Solution {
12 public:
13     ListNode* deleteDuplicates(ListNode* head) {
14         if(!head || !head->next){
15             return head;
16         }
17 
18         ListNode* dummy = new ListNode(-1, head);
19         ListNode* it = dummy;
20         while(it->next && it->next->next){
21             if(it->next->val != it->next->next->val){
22                 it = it->next;
23             }else{
24                 ListNode* dup = it->next->next;
25                 while(dup && dup->val == it->next->next->val){
26                     dup = dup->next;
27                 }
28 
29                 it->next = dup;
30             }
31         }
32 
33         return dummy->next;
34     }
35 };