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0876 - Middle of the Linked List (Easy)

https://leetcode.com/problems/middle-of-the-linked-list/description/

Problem Statement​

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

Approach 1: Fast and Slow Pointer​

Classic Fast and Slow Pointer question

  • Time Complexity: O(N)O(N) where NN is the number of nodes
  • Space Complexity: O(1)O(1)
Written by @wingkwong
/**
 * 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) {
        // fast & slow pointer
        // slow moves 1 step
        // fast moves 2 steps
        // 1 -> 2 -> 3 -> 4 -> 5
        // slow : 1 -> 2 -> 3
        // fast : 1 -> 3 -> 5
        // when fast reaches the end, slow will be the middle
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
};