Question

19. 删除链表的倒数第 N 个结点

English Version

题目描述

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

 

进阶：你能尝试使用一趟扫描实现吗？

解法

方法一：快慢指针

我们定义两个指针 fast 和 slow，初始时都指向链表的虚拟头结点 dummy。

接着 fast 指针先向前移动 $n$ 步，然后 fast 和 slow 指针同时向前移动，直到 fast 指针到达链表的末尾。此时 slow.next 指针指向的结点就是倒数第 n 个结点的前驱结点，将其删除即可。

时间复杂度 $O(n)$，其中 $n$ 为链表的长度。空间复杂度 $O(1)$。

# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
    dummy = ListNode(next=head)
    fast = slow = dummy
    for _ in range(n):
      fast = fast.next
    while fast.next:
      slow, fast = slow.next, 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, slow := dummy, dummy
  for ; n > 0; n-- {
    fast = fast.Next
  }
  for fast.Next != nil {
    slow, fast = slow.Next, fast.Next
  }
  slow.Next = slow.Next.Next
  return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
  const dummy = new ListNode(0, head);
  let fast = dummy;
  let 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.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//   ListNode {
//     next: None,
//     val
//   }
//   }
// }
impl Solution {
  pub fn remove_nth_from_end(head: Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> {
    let mut dummy = Some(Box::new(ListNode { val: 0, next: head }));
    let mut slow = &mut dummy;
    let mut fast = &slow.clone();
    for _ in 0..=n {
      fast = &fast.as_ref().unwrap().next;
    }
    while fast.is_some() {
      fast = &fast.as_ref().unwrap().next;
      slow = &mut slow.as_mut().unwrap().next;
    }
    slow.as_mut().unwrap().next = slow.as_mut().unwrap().next.as_mut().unwrap().next.take();
    dummy.unwrap().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

# Definition for singly-linked list.
# class ListNode {
#   public $val;
#   public $next;

#   public function __construct($val = 0, $next = null)
#   {
#     $this->val = $val;
#     $this->next = $next;
#   }
# }

class Solution {
  /**
   * @param ListNode $head
   * @param int $n
   * @return ListNode
   */

  function removeNthFromEnd($head, $n) {
    $dummy = new ListNode(0);
    $dummy->next = $head;

    $first = $dummy;
    $second = $dummy;

    for ($i = 0; $i <= $n; $i++) {
      $second = $second->next;
    }

    while ($second != null) {
      $first = $first->next;
      $second = $second->next;
    }

    $first->next = $first->next->next;

    return $dummy->next;
  }
}