LeetCode in Kotlin
2487. Remove Nodes From Linked List
Medium
You are given the head of a linked list.
Remove every node which has a node with a strictly greater value anywhere to the right side of it.
Return the head of the modified linked list.
Example 1:
Input: head = [5,2,13,3,8]
Output: [13,8]
Explanation: The nodes that should be removed are 5, 2 and 3.
- Node 13 is to the right of node 5.
- Node 13 is to the right of node 2.
- Node 8 is to the right of node 3.
Example 2:
Input: head = [1,1,1,1]
Output: [1,1,1,1]
Explanation: Every node has value 1, so no nodes are removed.
Constraints:
- The number of the nodes in the given 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
* }
*/
@Suppress("NAME_SHADOWING")
class Solution {
fun removeNodes(head: ListNode?): ListNode? {
var head = head
head = reverse(head)
if (head == null) {
return null
}
var max = head.`val`
var temp = head
var temp1 = head.next
while (temp1 != null) {
if (temp1.`val` >= max) {
max = temp1.`val`
temp!!.next = temp1
temp = temp.next
}
temp1 = temp1.next
}
temp!!.next = null
return reverse(head)
}
private fun reverse(head: ListNode?): ListNode? {
if (head == null || head.next == null) {
return head
}
var prev: ListNode? = null
var curr = head
var next: ListNode?
while (curr != null) {
next = curr.next
curr.next = prev
prev = curr
curr = next
}
return prev
}
}