LeetCode in Kotlin
2095. Delete the Middle Node of a Linked List
Medium
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
Solution
import com_github_leetcode.ListNode
/*
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
fun deleteMiddle(head: ListNode?): ListNode? {
var slow = head
var fast = head
var prev: ListNode? = null
while (fast?.next != null) {
prev = slow
slow = slow?.next
fast = fast.next?.next
}
if (slow == head) {
return null
}
prev?.next = slow?.next
return head
}
}