In this guide, you will get 876. Middle of the Linked List LeetCode Solution with the best time and space complexity. The solution to Middle of the Linked List problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
876. Middle of the Linked List LeetCode Solution in C++
class Solution {
public:
ListNode* middleNode(ListNode* head) {
ListNode* slow = head;
ListNode* fast = head;
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
}; /* code provided by PROGIEZ */
876. Middle of the Linked List LeetCode Solution in Java
class Solution {
public ListNode middleNode(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
} // code provided by PROGIEZ
876. Middle of the Linked List LeetCode Solution in Python
class Solution:
def middleNode(self, head: ListNode) -> ListNode:
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow # code by PROGIEZ
