LeetCode //C - 2095. Delete the Middle Node of a Linked List-Toy模板网

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2095. Delete the Middle Node of a Linked List

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:

LeetCode //C - 2095. Delete the Middle Node of a Linked List,LeetCode,leetcode,c语言,算法

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:

LeetCode //C - 2095. Delete the Middle Node of a Linked List,LeetCode,leetcode,c语言,算法

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:

LeetCode //C - 2095. Delete the Middle Node of a Linked List,LeetCode,leetcode,c语言,算法

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 $1, 10^5]$ .
$1 <= Node.val <= 10^5$

From: LeetCode
Link: 2095. Delete the Middle Node of a Linked List

Solution:

Ideas：

This code uses the fast and slow pointer technique to find the middle node. The fast pointer moves two steps at a time, while the slow pointer moves one step at a time. When the fast pointer reaches the end of the list, the slow pointer will be at the middle node. We then delete the middle node and return the modified list.文章来源地址https://www.toymoban.com/news/detail-783691.html

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* deleteMiddle(struct ListNode* head) {
    if (head == NULL || head->next == NULL) {
        free(head);
        return NULL;
    }
    
    struct ListNode *slow = head;
    struct ListNode *fast = head;
    struct ListNode *prev = NULL;
    
    // Use the fast and slow pointer approach to find the middle node
    while (fast != NULL && fast->next != NULL) {
        fast = fast->next->next;
        prev = slow;
        slow = slow->next;
    }
    
    // Delete the middle node
    prev->next = slow->next;
    free(slow);
    
    return head;
}