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.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode fast=head; ListNode slow=head; while(n-->0) { fast=fast.next; } if(fast==null)return head.next; while(fast.next!=null) { fast=fast.next; slow=slow.next; } slow.next=slow.next.next; return head; } }另一种方法 public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode start=new ListNode(0); start.next=head; ListNode fast=start; ListNode slow=start; for(;fast.next!=null;n--) { fast=fast.next; if(n<=0)slow=slow.next; } slow.next=slow.next.next; return start.next; } }