Thursday, June 15, 2017

19. Remove Nth Node From End of List

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.



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;
    }
}