Question

剑指 Offer II 021. 删除链表的倒数第 n 个结点

题目描述

给定一个链表，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

 

进阶：能尝试使用一趟扫描实现吗？

 

注意：本题与主站 19 题相同： https://leetcode.cn/problems/remove-nth-node-from-end-of-list/

解法

方法一

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
    dummy = ListNode(next=head)
    slow, fast = dummy, dummy
    for _ in range(n):
      fast = fast.next
    while fast.next:
      slow = slow.next
      fast = fast.next
    slow.next = slow.next.next
    return dummy.next

/**
 * 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, head);
    ListNode fast = dummy, slow = dummy;
    while (n-- > 0) {
      fast = fast.next;
    }
    while (fast.next != null) {
      slow = slow.next;
      fast = fast.next;
    }
    slow.next = slow.next.next;
    return dummy.next;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode() : val(0), next(nullptr) {}
 *   ListNode(int x) : val(x), next(nullptr) {}
 *   ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode* dummy = new ListNode(0, head);
    ListNode* fast = dummy;
    ListNode* slow = dummy;
    while (n--) {
      fast = fast->next;
    }
    while (fast->next) {
      slow = slow->next;
      fast = fast->next;
    }
    slow->next = slow->next->next;
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func removeNthFromEnd(head *ListNode, n int) *ListNode {
  dummy := &ListNode{0, head}
  fast := dummy
  slow := dummy
  for n > 0 {
    fast = fast.Next
    n -= 1
  }
  for fast.Next != nil {
    slow = slow.Next
    fast = fast.Next
  }
  slow.Next = slow.Next.Next
  return dummy.Next
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function (head, n) {
  const dummy = new ListNode(0, head);
  let fast = dummy,
    slow = dummy;
  while (n--) {
    fast = fast.next;
  }
  while (fast.next) {
    slow = slow.next;
    fast = fast.next;
  }
  slow.next = slow.next.next;
  return dummy.next;
};

# Definition for singly-linked list.
# class ListNode
#   attr_accessor :val, :next
#   def initialize(val = 0, _next = nil)
#     @val = val
#     @next = _next
#   end
# end
# @param {ListNode} head
# @param {Integer} n
# @return {ListNode}
def remove_nth_from_end(head, n)
  dummy = ListNode.new(0, head)
  fast = slow = dummy
  while n > 0
    fast = fast.next
    n -= 1
  end
  while fast.next
    slow = slow.next
    fast = fast.next
  end
  slow.next = slow.next.next
  return dummy.next
end