Given a linked list, remove the n th 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.
这题比较简单，使用快慢指针，找到倒数第n个结点的前驱即可，然后连接其前驱和后继即可，值得注意的是，两个while的条件之间的关系。代码如下：
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 { 13 ListNode *nList=new ListNode(-1); 14 nList->next=head; 15 ListNode *pre=nList; 16 ListNode *fast=head; 17 ListNode *slow=head; 18 int num=0; 19 20 while(num++<n) 21 { 22 fast=fast->next; 23 } 24 while(fast->next) 25 { 26 pre=pre->next; 27 slow=slow->next; 28 fast=fast->next; 29 } 30 pre->next=slow->next; 31 32 return nList->next; 33 } 34 };
[Leetcode] remove nth node from the end of list 删除链表倒数第n各节点
原文：http://www.cnblogs.com/love-yh/p/7050488.html