LeetCode in Kotlin
328. Odd Even Linked List
Medium
Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.
The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should remain as it was in the input.
You must solve the problem in O(1) extra space complexity and O(n) time complexity.
Example 1:
Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]
Example 2:
Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]
Constraints:
- The number of nodes in the linked list is in the range
[0, 104]. -106 <= Node.val <= 106
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 oddEvenList(head: ListNode?): ListNode? {
val odd = ListNode(0)
val even = ListNode(0)
var oddPointer = odd
var evenPointer = even
var pointer = head
var count = 0
while (pointer != null) {
if (count % 2 == 0) {
oddPointer.next = pointer
oddPointer = oddPointer.next!!
} else {
evenPointer.next = pointer
evenPointer = evenPointer.next!!
}
val next = pointer.next
pointer.next = null
pointer = next
count++
}
oddPointer.next = even.next
return odd.next
}
}