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:

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:

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:

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

Code

Time Complexity: $O(N)$, Space Complexity: $O(1)$

Direct Pointer

使用 Middle of the Linked List 中的快慢指標技巧，找到中點，接著再刪除中點即可。

注意 edge case：當只有一個 node 的時候。

/**
 * 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) {
            return nullptr;
        }
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* prev = nullptr;
        while(fast && fast->next) {
            prev = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        // now slow is the middle
        prev->next = slow->next;
 
        return head;
    }
};

Pointer to pointer

利用 pointer to pointer (indirect pointer) 的技巧，可以避免掉 edge case 的處理。

用 toBeFree 去保存要被刪掉的 node 之 address，如此一來就可以在使用 *indirect = (*indirect)->next; 去修改 address 後，將原本的 address free 掉，避免 memory leak。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteMiddle(struct ListNode* head) {
    struct ListNode** indirect = &head;
    struct ListNode* fast = head;
    while(fast && fast->next) {
        indirect = &(*indirect)->next;
        fast = fast->next->next;
    }
    struct ListNode* toBeFree = *indirect;
    *indirect = (*indirect)->next;
    free(toBeFree);
    return head;
}

/**
 * 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) {
        ListNode** slow = &head;
        ListNode* fast = head;
        while(fast && fast->next) {
            slow = &((*slow)->next);
            fast = fast->next->next;
        }
        
        *slow = (*slow)->next;
 
        return head;
    }
};

Source

• Delete the Middle Node of a Linked List - LeetCode