Question

2095. Delete the Middle Node of a Linked List

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Description

You are given the head of a linked list. Delete the middle node, and return _the_ head _of the modified linked list_.

The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x.

• For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively.

 

Example 1:

Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node. 

Example 2:

Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.

Example 3:

Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.

 

Constraints:

• The number of nodes in the list is in the range [1, 105].
• 1 <= Node.val <= 105

Solutions

Solution 1

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def deleteMiddle(self, head: Optional[ListNode]) -> Optional[ListNode]:
    dummy = ListNode(next=head)
    slow, fast = dummy, head
    while fast and fast.next:
      slow = slow.next
      fast = fast.next.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 deleteMiddle(ListNode head) {
    ListNode dummy = new ListNode(0, head);
    ListNode slow = dummy, fast = head;
    while (fast != null && fast.next != null) {
      slow = slow.next;
      fast = fast.next.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* deleteMiddle(ListNode* head) {
    ListNode* dummy = new ListNode(0, head);
    ListNode* slow = dummy;
    ListNode* fast = head;
    while (fast && fast->next) {
      slow = slow->next;
      fast = fast->next->next;
    }
    slow->next = slow->next->next;
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func deleteMiddle(head *ListNode) *ListNode {
  dummy := &ListNode{Val: 0, Next: head}
  slow, fast := dummy, dummy.Next
  for fast != nil && fast.Next != nil {
    slow, fast = slow.Next, fast.Next.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 deleteMiddle(head: ListNode | null): ListNode | null {
  if (!head || !head.next) return null;
  let fast = head.next,
    slow = head;
  while (fast.next && fast.next.next) {
    slow = slow.next;
    fast = fast.next.next;
  }
  slow.next = slow.next.next;
  return head;
}