19. 删除链表的倒数第N个节点 

 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 
13  //双指针算法
14  //删除->都会建立虚拟头节点
15 class Solution 
16 {
17 public:
18     ListNode* removeNthFromEnd(ListNode* head, int n) 
19     {
20         ListNode* dummy = new ListNode(-1);
21         dummy->next = head;
22 
23         ListNode* fast = dummy;
24         ListNode* slow = dummy;
25         for(int i = 0;i <= n;i ++) fast = fast->next;
26         while(fast)
27         {
28             fast = fast->next;
29             slow = slow->next;
30         }
31         slow->next = slow->next->next;//不需要判空，因为有虚拟头节点的存在
32         return dummy->next; 
33     }
34 };

 

posted @ 2020-03-15 18:34  Jinxiaobo0509  阅读(100)  评论(0)    收藏  举报