LeetCode 0876. Middle of the Linked List Solution in Java, C++, Python & More | Explanation + Code

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0876. Middle of the Linked List

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: Fast and Slow Pointers

We define two pointers fast and slow, both initially pointing to the head of the linked list.

The fast pointer fast moves two steps at a time, while the slow pointer slow moves one step at a time. When the fast pointer reaches the end of the linked list, the node pointed to by the slow pointer is the middle node.

The time complexity is O(n), where n is the length of the linked list. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustPHPC
# 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(code-box)

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