[LeetCode19]Remove Nth Node From End of List

Posted 2020-06-13

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

给定一个单链表，移除倒数第n个Node，返回链表的head

思路：维护两个指针slow，fast，fast先移动n步，然后slow、fast同时移动，直到fast为空为止，删除slow后面的结点即可

代码：

 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 class Solution {
10 public:
11     ListNode* removeNthFromEnd(ListNode* head, int n) {
12         ListNode *preHead = new ListNode(0);
13         ListNode *fast = preHead, *slow = preHead;
14         slow->next = head;
15         for(int i = 0; i <= n; ++i)
16             fast = fast->next;
17         while(fast)
18         {
19             fast = fast->next;
20             slow = slow->next;
21         }
22         slow->next = slow->next->next;
23         return preHead->next;
24     }
25 };