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
, and5
, the middle nodes are0
,1
,1
,2
, and2
, 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
Similar Questions:
- Remove Nth Node From End of List (Medium)
- Reorder List (Medium)
- Remove Linked List Elements (Easy)
- Middle of the Linked List (Easy)
// OJ: https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
int getLength(ListNode *head) {
int ans = 0;
for (; head; ++ans, head = head->next);
return ans;
}
public:
ListNode* deleteMiddle(ListNode* head) {
ListNode dummy, *p = &dummy;
dummy.next = head;
int len = getLength(head) / 2;
for (int i = 0; i < len; ++i, p = p->next);
p->next = p->next->next;
return dummy.next;
}
};
// OJ: https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode* deleteMiddle(ListNode* head) {
ListNode dummy, *fast = &dummy, *slow = &dummy;
dummy.next = head;
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
}
slow->next = slow->next->next;
return dummy.next;
}
};