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,

1
2
3

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

 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class  {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode first = head, second = head, pre = null;
        while((--n)>0&&first.next!=null){
            first = first.next;
        }
        while(first.next!=null) {
            first = first.next;
            pre = second;
            second = second.next;
        }
        if(second.next==null&&pre==null) {
            head = null;
        } else if(second.next==null) {
            pre.next = null;
        } else if(pre == null) {
            head = head.next;
        } else {
            pre.next = second.next;
        }
        return head;
    }
}

remove nth node from end of list

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