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.

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
 
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode faster = head;
	    ListNode slower = head;
	    while (n > 0 && faster != null) {
	        faster = faster.next;
	        n--;
	    }
	    // Check if has only node
	    if (faster == null) return head.next; 

	    while (faster.next != null) {
	        faster = faster.next;
	        slower = slower.next;
	    }
	    // Remove slower.next which is the nth form the end
	    slower.next = slower.next.next;
	    return head;
    }
}