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.
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode removeNthFromEnd(ListNode head, int n) { 14 ArrayList<ListNode> list=new ArrayList<>(); 15 ListNode node=head; 16 int index=0; 17 while (node != null) { 18 list.add(node); 19 node=node.next; 20 } 21 index=list.size()-n; 22 if (index ==0) { 23 return head.next; 24 } 25 if (index==list.size()-1) { 26 list.get(index-1).next=null; 27 return head; 28 } 29 30 list.get(index-1).next=list.get(index+1); 31 return head; 32 33 } 34 }
LeetCode Remove Nth Node From End of List
原文：http://www.cnblogs.com/birdhack/p/4047919.html