leetcode-双指针-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="width: 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>
 */

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/**
 * 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;
    }
}
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