LeetCode-in-All
148. Sort List
Medium
Given the head of a linked list, return the list after sorting it in ascending order.
Example 1:
Input: head = [4,2,1,3]
Output: [1,2,3,4]
Example 2:
Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
Example 3:
Input: head = []
Output: []
Constraints:
- The number of nodes in the list is in the range
[0, 5 * 104]. -105 <= Node.val <= 105
Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?
Solution
import com_github_leetcode.ListNode
/*
* Definition for singly-linked list.
* class ListNode(_x: Int = 0, _next: ListNode = null) {
* var next: ListNode = _next
* var x: Int = _x
* }
*/
object Solution {
def sortList(head: ListNode): ListNode = {
if (head == null || head.next == null) {
return head
}
val (first, second) = splitList(head)
mergeLists(sortList(first), sortList(second))
}
def splitList(head: ListNode): (ListNode, ListNode) = {
var slow = head
var fast = head
var pre = slow
while (fast != null && fast.next != null) {
pre = slow
slow = slow.next
fast = fast.next.next
}
pre.next = null
(head, slow)
}
private def mergeLists(first: ListNode, second: ListNode): ListNode = {
val res = new ListNode(0)
var cur = res
var (firstPtr, secondPtr) = (first, second)
while (firstPtr != null || secondPtr != null) {
if (firstPtr != null && (secondPtr == null || firstPtr.x <= secondPtr.x)) {
cur.next = firstPtr
firstPtr = firstPtr.next
} else {
cur.next = secondPtr
secondPtr = secondPtr.next
}
cur = cur.next
}
res.next
}
}