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2046. Sort Linked List Already Sorted Using Absolute Values

Description

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?

Solutions

  • /**
     * 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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