LeetCode:Remove Nth Node From End of List

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.

Note:
Given n will always be valid.
Try to do this in one pass.

主要难点是通过一趟遍历寻找链表倒数第k个元素，具体见代码注释本文地址

 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         //快指针先走n步，然后快慢指针一起走，块指针指向尾节点时，慢指针指向倒数第n个节点
13         ListNode* fast = head, *slow = head, *slowpre = NULL;
14         for(int i = 1; i < n; i++)fast = fast->next;
15         while(fast->next)
16         {
17             fast = fast->next;
18             slowpre = slow;
19             slow = slow->next;
20         }
21         if(slow == head)
22             head = head->next;
23         else
24             slowpre->next = slow->next;
25         delete slow;
26         return head;
27     }
28 };

原文地址：https://www.cnblogs.com/TenosDoIt/p/3667510.html