#yyds干货盘点# leetcode链表-删除链表的倒数第 n 个结点

删除链表的倒数第n各节点难点再入如何找到倒数第n个节点。在不使用额外空间的情况下，可以通过计算或者通过双指针去寻找，这里有俩种方法；第一种 计算，第二种 双指针,效率上第二种比较快

/**
 * <p>给你一个链表，删除链表的倒数第 <code>n</code><em> </em>个结点，并且返回链表的头结点。</p>
 *
 * <p> </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> </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 <= sz <= 30</code></li>
 * <li><code>0 <= Node.val <= 100</code></li>
 * <li><code>1 <= n <= sz</code></li>
 * </ul>
 *
 * <p> </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(0);
        dummy.next= head;

        ListNode countP = head;
        int len =0;
        while(countP!=null){
            len++;
            countP=countP.next;
        }


        System.out.println("len " +len);

        int currIndex = 0;
        int removeIndex =  len+1-n-1;
        System.out.println("removeIndex " +removeIndex);

        ListNode curr = dummy;
        while(curr!=null){
            if(currIndex==removeIndex){
                curr.next=curr.next.next;
                return dummy.next;
            }else{
                currIndex++;
                curr=curr.next;
            }
        }
        return dummy.next;
    }
}
//leetcode submit region end(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;
  }
}