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 = 
Output: 
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.

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

Related Topics:
Linked List, Two Pointers, Sorting

Similar Questions:

## Solution 1.

// OJ: https://leetcode.com/problems/sort-linked-list-already-sorted-using-absolute-values/
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode dummy;
auto node = head->next;
if (node->val >= 0) {
} else {
node->next = dummy.next;
dummy.next = node;
}
}
return dummy.next;
}
};


## Solution 2.

// OJ: https://leetcode.com/problems/sort-linked-list-already-sorted-using-absolute-values/
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode pos, *posTail = &pos, neg, *negTail = NULL;
auto node = head;
if (node->val >= 0) {
posTail->next = node;
posTail = node;
} else {
node->next = neg.next;
neg.next = node;
if (!negTail) negTail = node;
}
}
if (negTail) {
negTail->next = pos.next;
posTail->next = NULL;
return neg.next;
}
return pos.next;
}
};