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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:
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* sortLinkedList(ListNode* head) {
ListNode dummy;
dummy.next = head;
while (head->next) {
auto node = head->next;
if (node->val >= 0) {
head = head->next;
} else {
head->next = node->next;
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* sortLinkedList(ListNode* head) {
ListNode pos, *posTail = &pos, neg, *negTail = NULL;
while (head) {
auto node = head;
head = head->next;
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;
}
};