LeetCode //C - 237. Delete Node in a Linked List

237. Delete Node in a Linked List

There is a singly-linked list head and we want to delete a node node in it.

You are given the node to be deleted node. You will not be given access to the first node of head.

All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list.

Delete the given node. Note that by deleting the node, we do not mean removing it from memory. We mean:

The value of the given node should not exist in the linked list.
The number of nodes in the linked list should decrease by one.
All the values before node should be in the same order.
All the values after node should be in the same order.

Custom testing:

For the input, you should provide the entire linked list head and the node to be given node. node should not be the last node of the list and should be an actual node in the list.
We will build the linked list and pass the node to your function.
The output will be the entire list after calling your function.

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Constraints:

The number of the nodes in the given list is in the range [2, 1000].
-1000 <= Node.val <= 1000
The value of each node in the list is unique.
The node to be deleted is in the list and is not a tail node.

From: LeetCode

Link: 237. Delete Node in a Linked List

Solution:

Ideas:

Check for edge cases: The function first checks if the node to be deleted is NULL or if it is the last node. If either condition is true, the function returns immediately since it is not possible to delete the last node or a NULL node with this approach.
Copy the next node's value: The value of the next node is copied to the given node.
Link to the next of next node: The next pointer of the given node is updated to skip the next node, effectively linking to the node after the next node.
Free the next node: The memory of the next node (which is now redundant) is freed.

Code:

c 复制代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void deleteNode(struct ListNode* node) {
    if (node == NULL || node->next == NULL) {
        return; // Cannot delete if the node is NULL or it is the last node
    }
    
    struct ListNode* nextNode = node->next;
    node->val = nextNode->val;
    node->next = nextNode->next;
    free(nextNode); // Free the memory of the next node
}