0100 - 0199
0148 - Sort List (Medium)
Problem Link
https://leetcode.com/problems/sort-list/
Problem Statement
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:
- The number of nodes in the list is in the range
[0, 5 * 10^4]. -10^5 <= Node.val <= 10^5
Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?
Approach 1: Recursive Merge Sort
- Find the middle node and cut the head reference till middle node
- Keep reducing the nodes size to smaller for comparison (same as like merge sort)
- Once we reduce nodes size to 1, merge the nodes in sorted (ascending) order.
- Keep merging the nodes till last, to build the sorted list.
Time Complexity: , where - # of nodes in the list
Space complexity: , - recursive call stack
/**
* 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 sortList(ListNode head) {
if (Objects.isNull(head) || Objects.isNull(head.next)) {
return head;
}
// Middle node
ListNode mid = middleNode(head);
// Keep traversing left to get the smallest nodes for comparison (smallest we can get is 1 node)
ListNode left = sortList(head);
// Starting from middle, to find the smallest nodes for comparison
ListNode right = sortList(mid);
// Compare the list and return the merged nodes
return mergeTwoLists(left, right);
}
public ListNode middleNode(ListNode head) {
ListNode midPrev = null;
//
while (head != null && head.next != null) {
midPrev = (midPrev == null) ? head : midPrev.next;
head = head.next.next;
}
ListNode mid = midPrev.next;
// Cut the reference to the next pointer (mid), so that head remains from start to mid.
midPrev.next = null;
return mid;
}
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
// Base case
if (Objects.isNull(list1) && Objects.isNull(list2)) {
return list1;
}
if (Objects.isNull(list1)) {
return list2;
}
if (Objects.isNull(list2)) {
return list1;
}
ListNode head = new ListNode();
ListNode node = head;
while (Objects.nonNull(list1) && Objects.nonNull(list2)) {
if (list1.val <= list2.val) {
node.next = list1;
list1 = list1.next;
} else {
node.next = list2;
list2 = list2.next;
}
node = node.next;
}
// If either of half is not empty then append it
node.next = Objects.nonNull(list1) ? list1 : list2;
return head.next;
}
}