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Formatted question description: https://leetcode.ca/all/2046.html

2046. Sort Linked List Already Sorted Using Absolute Values (Medium)

Given the head of a singly linked list that is sorted in non-decreasing order using the absolute values of its nodes, return the list sorted in non-decreasing order using the actual values of its nodes.

 

Example 1:

Input: head = [0,2,-5,5,10,-10]
Output: [-10,-5,0,2,5,10]
Explanation:
The list sorted in non-descending order using the absolute values of the nodes is [0,2,-5,5,10,-10].
The list sorted in non-descending order using the actual values is [-10,-5,0,2,5,10].

Example 2:

Input: head = [0,1,2]
Output: [0,1,2]
Explanation:
The linked list is already sorted in non-decreasing order.

Example 3:

Input: head = [1]
Output: [1]
Explanation:
The linked list is already sorted in non-decreasing order.

 

Constraints:

• The number of nodes in the list is the range [1, 105].
• -5000 <= Node.val <= 5000
• head is sorted in non-decreasing order using the absolute value of its nodes.

 

Follow up:

• Can you think of a solution with O(n) time complexity?

Related Topics:
Linked List, Two Pointers, Sorting

Similar Questions:

• Sort List (Medium)

Solution 1.

• Java
• C++
• Python
• Go

• /**
   * Definition for singly-linked list.
   * public class ListNode {
   *     int val;
   *     ListNode next;
   *     ListNode() {}
   *     ListNode(int val) { this.val = val; }
   *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
   * }
   */
  class Solution {
      public ListNode sortLinkedList(ListNode head) {
          ListNode prev = head, curr = head.next;
          while (curr != null) {
              if (curr.val < 0) {
                  ListNode t = curr.next;
                  prev.next = t;
                  curr.next = head;
                  head = curr;
                  curr = t;
              } else {
                  prev = curr;
                  curr = curr.next;
              }
          }
          return head;
      }
  }
  

• /**
   * Definition for singly-linked list.
   * struct ListNode {
   *     int val;
   *     ListNode *next;
   *     ListNode() : val(0), next(nullptr) {}
   *     ListNode(int x) : val(x), next(nullptr) {}
   *     ListNode(int x, ListNode *next) : val(x), next(next) {}
   * };
   */
  class Solution {
  public:
      ListNode* sortLinkedList(ListNode* head) {
          ListNode* prev = head;
          ListNode* curr = head->next;
          while (curr) {
              if (curr->val < 0) {
                  auto t = curr->next;
                  prev->next = t;
                  curr->next = head;
                  head = curr;
                  curr = t;
              } else {
                  prev = curr;
                  curr = curr->next;
              }
          }
          return head;
      }
  };
  

• # Definition for singly-linked list.
  # class ListNode:
  #     def __init__(self, val=0, next=None):
  #         self.val = val
  #         self.next = next
  class Solution:
      def sortLinkedList(self, head: Optional[ListNode]) -> Optional[ListNode]:
          prev, curr = head, head.next
          while curr:
              if curr.val < 0:
                  t = curr.next
                  prev.next = t
                  curr.next = head
                  head = curr
                  curr = t
              else:
                  prev, curr = curr, curr.next
          return head
  
  

• /**
   * Definition for singly-linked list.
   * type ListNode struct {
   *     Val int
   *     Next *ListNode
   * }
   */
  func sortLinkedList(head *ListNode) *ListNode {
  	prev, curr := head, head.Next
  	for curr != nil {
  		if curr.Val < 0 {
  			t := curr.Next
  			prev.Next = t
  			curr.Next = head
  			head = curr
  			curr = t
  		} else {
  			prev, curr = curr, curr.Next
  		}
  	}
  	return head
  }
  

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