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:

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;    }}