LeetCode 75 - 2095. Delete the Middle Node of a Linked List

Description

You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list.

The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x.

For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively.

Example 1:

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Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node. 

Example 2:

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Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.

Example 3:

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Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.

Constraints:

The number of nodes in the list is in the range [1, 105].
1 <= Node.val <= 105

Solutions

Fast and Slow Pointer Approach: Use two pointers, fast and slow, to traverse the linked list. The fast pointer moves two steps at a time, while the slow pointer moves one step at a time. By the time the fast pointer reaches the end of the list, the slow pointer will be at the middle of the list.

Note: If the linked list has only one node, we remove this node and return an empty list. However, since this node does not have a previous node, current is meaningless. Therefore, we can make a special check before traversal and directly return a null pointer as the answer.

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/**
 * 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* deleteMiddle(ListNode* head) {
        if (head->next == nullptr) {
            return nullptr;
        }

        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* current{};

        while (fast != nullptr && fast->next != nullptr) {
            fast = fast->next->next;
            current = slow;
            slow = slow->next;
        }

        current->next = current->next->next;

        return head;
    }
};

Why does the slow pointer end up at the midpoint?

Let the length of the linked list be ( n ) (with node indices starting from 0, where the head node has index 0). The slow pointer moves at a speed of 1 node per step, while the fast pointer moves at a speed of 2 nodes per step. After ( k ) iterations (i.e., after the slow pointer has moved ( k ) steps), the slow pointer is at index ( k ), and the fast pointer is at index ( 2k ).