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:
[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)?
import { ListNode } from '../../com_github_leetcode/listnode'
/*
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function sortList(head: ListNode | null): ListNode | null {
if (!head) return null
let array = []
while (head) {
array.push([head, head.val])
head = head.next
}
array.sort((a, b) => {
return a[1] - b[1]
})
for (let iter = 0; iter < array.length; iter++) {
if (iter + 1 >= array.length) array[iter][0].next = null
else array[iter][0].next = array[iter + 1][0]
}
return array[0][0]
}
export { sortList }