LeetCode in Kotlin
86. Partition List
Medium
Given the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
Example 1:
Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]
Example 2:
Input: head = [2,1], x = 2
Output: [1,2]
Constraints:
- The number of nodes in the list is in the range
[0, 200]. -100 <= Node.val <= 100-200 <= x <= 200
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 partition(head: ListNode?, x: Int): ListNode? {
var head = head
var nHead: ListNode? = ListNode(0)
var nTail: ListNode? = ListNode(0)
val ptr = nTail
val temp = nHead
while (head != null) {
val nNext = head.next
if (head.`val` < x) {
nHead!!.next = head
nHead = nHead.next
} else {
nTail!!.next = head
nTail = nTail.next
}
head = nNext
}
nTail!!.next = null
nHead!!.next = ptr!!.next
return temp!!.next
}
}