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
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
| # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def deleteMiddle(self, head: Optional[ListNode]) -> Optional[ListNode]:
dummy = ListNode(next=head)
slow, fast = dummy, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
slow.next = slow.next.next
return dummy.next
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
| /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteMiddle(ListNode head) {
ListNode dummy = new ListNode(0, head);
ListNode slow = dummy, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
slow.next = slow.next.next;
return dummy.next;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
| /**
* 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* dummy = new ListNode(0, head);
ListNode* slow = dummy;
ListNode* fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
slow->next = slow->next->next;
return dummy->next;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
| /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func deleteMiddle(head *ListNode) *ListNode {
dummy := &ListNode{Val: 0, Next: head}
slow, fast := dummy, dummy.Next
for fast != nil && fast.Next != nil {
slow, fast = slow.Next, fast.Next.Next
}
slow.Next = slow.Next.Next
return dummy.Next
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
| /**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function deleteMiddle(head: ListNode | null): ListNode | null {
if (!head || !head.next) return null;
let fast = head.next,
slow = head;
while (fast.next && fast.next.next) {
slow = slow.next;
fast = fast.next.next;
}
slow.next = slow.next.next;
return head;
}
|
