Question

876. Middle of the Linked List

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Description

Given the head of a singly linked list, return _the middle node of the linked list_.

If there are two middle nodes, return the second middle node.

 

Example 1:

Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.

Example 2:

Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.

 

Constraints:

• The number of nodes in the list is in the range [1, 100].
• 1 <= Node.val <= 100

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 middleNode(self, head: ListNode) -> ListNode:
    slow = fast = head
    while fast and fast.next:
      slow, fast = slow.next, fast.next.next
    return slow

/**
 * 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 middleNode(ListNode head) {
    ListNode slow = head, fast = head;
    while (fast != null && fast.next != null) {
      slow = slow.next;
      fast = fast.next.next;
    }
    return slow;
  }
}

/**
 * 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* middleNode(ListNode* head) {
    ListNode *slow = head, *fast = head;
    while (fast && fast->next) {
      slow = slow->next;
      fast = fast->next->next;
    }
    return slow;
  }
};

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

/**
 * 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 middleNode(head: ListNode | null): ListNode | null {
  let fast = head,
    slow = head;
  while (fast != null && fast.next != null) {
    fast = fast.next.next;
    slow = slow.next;
  }
  return slow;
}

// 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 middle_node(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    let mut slow = &head;
    let mut fast = &head;
    while fast.is_some() && fast.as_ref().unwrap().next.is_some() {
      slow = &slow.as_ref().unwrap().next;
      fast = &fast.as_ref().unwrap().next.as_ref().unwrap().next;
    }
    slow.clone()
  }
}

/**
 * Definition for a singly-linked list.
 * class ListNode {
 *   public $val = 0;
 *   public $next = null;
 *   function __construct($val = 0, $next = null) {
 *     $this->val = $val;
 *     $this->next = $next;
 *   }
 * }
 */
class Solution {
  /**
   * @param ListNode $head
   * @return ListNode
   */
  function middleNode($head) {
    $count = 0;
    $tmpHead = $head;
    while ($tmpHead != null) {
      $tmpHead = $tmpHead->next;
      $count++;
    }
    $len = $count - floor($count / 2);
    while ($count != $len) {
      $head = $head->next;
      $count--;
    }
    return $head;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   struct ListNode *next;
 * };
 */

struct ListNode* middleNode(struct ListNode* head) {
  struct ListNode* fast = head;
  struct ListNode* slow = head;
  while (fast && fast->next) {
    fast = fast->next->next;
    slow = slow->next;
  }
  return slow;
}