【朱迪的LeetCode刷题笔记】82. Remove Duplicates from Sorted List II #Medium #C

Judie

1. 
   
2. 82. Remove Duplicates from Sorted List II #Medium
   
3. 
   
4. Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.
   
5. 
   
6. 
   
7. Example 1:
   
8. 
   
9. Input: head = [1,2,3,3,4,4,5]
   
10. Output: [1,2,5]
    
11. 
    
12. 
    
13. Example 2:
    
14. 
    
15. Input: head = [1,1,1,2,3]
    
16. Output: [2,3]
    
17. 
    
18. 
    
19. Constraints:
    
20. 
    
21. The number of nodes in the list is in the range [0, 300].
    
22. -100 <= Node.val <= 100
    
23. The list is guaranteed to be sorted in ascending order.
    

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C

1. 
   
2. /**
   
3. * Definition for singly-linked list.
   
4. * struct ListNode {
   
5. *     int val;
   
6. *     struct ListNode *next;
   
7. * };
   
8. */
   
9. 
   
10. 
    
11. struct ListNode* deleteDuplicates(struct ListNode* head) {
    
12.     struct ListNode *before = NULL;
    
13.     struct ListNode *current = head;
    
14.     if (head == NULL) {
    
15.         return NULL;
    
16.     }
    
17.     struct ListNode *next = head->next;
    
18.     while (next) {
    
19.         if (current->val == next->val) {
    
20.             while (next && current->val == next->val) {
    
21.                 current = current->next;
    
22.                 next = next->next;
    
23.             }
    
24.             if (next == NULL) {
    
25.                 if (before == NULL) {
    
26.                     return NULL;
    
27.                 }
    
28.                 before->next = NULL;
    
29.                 return head;
    
30.             }
    
31.             current = current->next;
    
32.             next = next->next;
    
33.             if (before == NULL) {
    
34.                 head = current;
    
35.             } else {
    
36.                 before->next = current;
    
37.             }
    
38.         } else {
    
39.             before = current;
    
40.             current = current->next;
    
41.             next = next->next;
    
42.         }     
    
43.     }
    
44.     return head;
    
45. }
    

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不愧是medium 想了三天qwq