19.删除链表的倒数第N个节点——学习笔记

题目：给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

提示：

• 链表中结点的数目为 sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

题目来源：力扣（LeetCode）链接

题解：

• 自己做的

  /**
  * 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) {
          int size = 0;//首先遍历链表，记录链表的总长度
          ListNode countNode = head;
          while (countNode != null) {
              size++;
              countNode = countNode.next;
          }
          //定义一个虚拟头节点
          ListNode dummyNode = new ListNode(-1);
          dummyNode.next = head;
          //查找倒数第n个节点的前一个节点
          ListNode prevNode = dummyNode;
          for (int i = 0; i < size - n; i++) {
              prevNode = prevNode.next;
          }
          //删除该节点
          prevNode.next = prevNode.next.next;
          //返回链表的真实头节点
          return dummyNode.next;
      }
  }
  

• 双指针法

  /**
  * 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) {
          //定义虚拟头节点
          ListNode dummyNode = new ListNode(-1);
          dummyNode.next = head;
          //定义快指针和慢指针
          ListNode fastNode = dummyNode;
          ListNode slowNode = dummyNode;
          //让快指针前移n个，保证其与慢指针之间相隔n个节点
          for (int i = 0; i < n; i++) {
              fastNode = fastNode.next;
          }
          //快指针指向最后一个节点时，慢指针正好指向倒数第n个节点的前一节点，因为快慢指针之间始终相隔n个节点
          while (fastNode.next != null) {
              fastNode = fastNode.next;
              slowNode = slowNode.next;
          }
          //删除倒数第n个节点
          slowNode.next = slowNode.next.next;
          //返回链表的真实头节点
          return dummyNode.next;
      }
  }