notes/OJ notes/pages/Leetcode Middle-of-the-Linked-List.md

Leetcode Middle-of-the-Linked-List

2022-07-13 09:08

Algorithms:

#algorithm #two_pointers

Data structures:

#DS #linked_list

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#two_pointers

• Leetcode Intersection-of-Two-Arrays-II
• Leetcode Merge-Sorted-Array
• Leetcode Merge-Two-Sorted-Lists
• Leetcode Move-Zeroes
• Leetcode Remove-Nth-Node-From-End-of-List
• Leetcode Reverse-Words-In-a-String
• Leetcode Squares-of-a-Sorted-Array
• Leetcode Two-Sum-II-Input-Array-Is-Sorted
• Two pointers approach

Links:

• Link to problem

---

Problem

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.

Examples

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

Thoughts

[!summary] This is a #two_pointers problem To find the middle of linked list, use two pointers: one fast pointer and one small pointer

Solution

/**
 * 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) {
    // two pointers problem.
    ListNode *fast, *slow;
    fast = slow = head;

    while (true) {
      if (fast->next && fast->next->next) {
        slow = slow->next;
        fast = fast->next->next;
      } else if (!fast->next) {
        return slow;
      } else if (!fast->next->next) {
        return slow->next;
      }
    }
  }
};