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.
tag: two pointers, dummy node. 尽量少的指针完成。
链表基础: reverse linked list
/** * 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 dummy = new ListNode(0); dummy.next = head; ListNode preN = dummy; for(int i = 0; i < n; i++) { if(head == null) { return null; } head = head.next; } while(head != null) { head = head.next; preN = preN.next; } preN.next = preN.next.next; return dummy.next; } }