給定一個連結串列，刪除連結串列中倒數第n個節點，返回連結串列的頭節點。
Given a linked list, remove the nth node from the end of list and return its head.

樣例
給出連結串列1->2->3->4->5->null和 n = 2.
刪除倒數第二個節點之後，這個連結串列將變成1->2->3->5->null.
Example
Given linked list: 1->2->3->4->5->null, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5->null.

注意
連結串列中的節點個數大於等於n
Note
The minimum number of nodes in list is n.

挑戰
O(n)時間複雜度
Challenge
O(n) time

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: The head of linked list.
     */
    ListNode removeNthFromEnd(ListNode head, int n) {
        if(n == 0)return head;
        ListNode p = head, q = head;
        while (n > 0
 
 && null != p.next) {
            p = p.next;
            n--;
        }
        if(n > 0) {
            head = head.next;
            return head;
        }
        while(null != p.next) {
            p = p.next;
            q = q.next;
        }
        q.next = q.next.next;
        return head;
    }
}