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.
Given n will always be valid.
Try to do this in one pass.
Solution:
We create a fast node and a slow node at the dummy node position.
Fast node advances n steps first.
Then both nodes advance together until fast node's next is null.
Now slow node's next node is the one to be removed.
The time complexity is O(n).
Code:
/** * 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 fast = dummy; ListNode slow = dummy; for (int i = 0; i < n; i++) { fast = fast.next; } while (fast != null && fast.next != null) { fast = fast.next; slow = slow.next; } slow.next = slow.next.next; return dummy.next; } }