LeetCode 148: Sort List Solution

Master LeetCode problem 148 (Sort List), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

148. Sort List

Medium

Linked List Two Pointers Divide and Conquer Sorting Merge Sort

Problem Explanation

Explanation

To sort a linked list in O(n logn) time and O(1) space, we can use the merge sort algorithm. The idea is to divide the list into two halves, sort each half recursively, and then merge the sorted halves back together. This approach guarantees the desired time complexity and uses only a constant amount of extra space.

Splitting: Find the middle of the linked list using the slow and fast pointer technique.
Recursively sort: Sort the two halves of the list recursively.
Merge: Merge the two sorted halves back together.

Time complexity: O(n logn)
Space complexity: O(1)

Solution Code

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode mid = findMiddle(head);
        ListNode right = mid.next;
        mid.next = null; // break the list
        
        ListNode leftSorted = sortList(head);
        ListNode rightSorted = sortList(right);
        
        return merge(leftSorted, rightSorted);
    }
    
    private ListNode findMiddle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                current.next = l1;
                l1 = l1.next;
            } else {
                current.next = l2;
                l2 = l2.next;
            }
            current = current.next;
        }
        
        if (l1 != null) {
            current.next = l1;
        } else {
            current.next = l2;
        }
        
        return dummy.next;
    }
}

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