leetcode 单链表19

/**
<p>给你一个链表，删除链表的倒数第&nbsp;<code>n</code><em>&nbsp;</em>个结点，并且返回链表的头结点。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style=" 542px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2,3,4,5], n = 2
<strong>输出：</strong>[1,2,3,5]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>head = [1], n = 1
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>head = [1,2], n = 1
<strong>输出：</strong>[1]
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中结点的数目为 <code>sz</code></li>
	<li><code>1 &lt;= sz &lt;= 30</code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= sz</code></li>
</ul>

<p>&nbsp;</p>

<p><strong>进阶：</strong>你能尝试使用一趟扫描实现吗？</p>
<div><div>Related Topics</div><div><li>链表</li><li>双指针</li></div></div><br><div><li> 1953</li><li> 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
		ListNode dummy = new ListNode(-1);
		dummy.next = head;
		ListNode p = findFromEnd(dummy,n+1);
		p.next = p.next.next;
		return dummy.next;
    }

	/**
	 * 5个节点，倒数第三个既第2个节点；要删除2就要找到第1个节点
	 * 5-3=2，传入的虚拟节点既 6-3 = 3;是原先链表的第二节点
	 * 
	 * @param head
	 * @param k
	 * @return
	 */
	ListNode findFromEnd(ListNode head,int k){
		ListNode p1 = head;
		for (int i = 0; i <k; i++) {
			p1 = p1.next;
		}
		ListNode p2= head;
		while(p1!=null){
			p1=p1.next;
			p2=p2.next;
		}
		return p2;
	}
}
//leetcode submit region end(Prohibit modification and deletion)