19. 删除链表的倒数第N个节点

Posted 2021-03-05

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 
10  //1、删除节点，就是找到该节点的前一个节点就可以
11  //2、然后slow->next = slow->next->next;
12 class Solution 
13 {
14 public:
15     ListNode* removeNthFromEnd(ListNode* head, int n) 
16     {
17         ListNode* dummy = new ListNode(-1);
18         dummy->next = head;
19 
20         ListNode* fast = dummy;
21         ListNode* slow = dummy;
22         for(int i = 0;i <= n;i ++) fast = fast->next;
23         while(fast)
24         {
25             fast = fast->next;
26             slow = slow->next;
27         }
28         if(slow->next == NULL) slow = NULL;
29         else slow->next = slow->next->next;
30         return dummy->next; 
31     }
32 };