LeetCode #148 — MEDIUM

Sort List

Move from brute-force thinking to an efficient approach using linked list strategy.

The Problem

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 * 10⁴].
-10⁵ <= Node.val <= 10⁵

Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?

Step 01

Brute Force Baseline

Problem summary: Given the head of a linked list, return the list after sorting it in ascending order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List · Two Pointers

Example 1

[4,2,1,3]

Example 2

[-1,5,3,4,0]

Example 3

[]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #148: Sort List
/**
 * 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 (head == null || head.next == null) {
            return head;
        }
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode l1 = head, l2 = slow.next;
        slow.next = null;
        l1 = sortList(l1);
        l2 = sortList(l2);
        ListNode dummy = new ListNode();
        ListNode tail = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                tail.next = l1;
                l1 = l1.next;
            } else {
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;
        }
        tail.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}

// Accepted solution for LeetCode #148: Sort List
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func sortList(head *ListNode) *ListNode {
	if head == nil || head.Next == nil {
		return head
	}
	slow, fast := head, head.Next
	for fast != nil && fast.Next != nil {
		slow, fast = slow.Next, fast.Next.Next
	}
	l1 := head
	l2 := slow.Next
	slow.Next = nil
	l1 = sortList(l1)
	l2 = sortList(l2)
	dummy := &ListNode{}
	tail := dummy
	for l1 != nil && l2 != nil {
		if l1.Val <= l2.Val {
			tail.Next = l1
			l1 = l1.Next
		} else {
			tail.Next = l2
			l2 = l2.Next
		}
		tail = tail.Next
	}
	if l1 != nil {
		tail.Next = l1
	} else {
		tail.Next = l2
	}
	return dummy.Next
}

# Accepted solution for LeetCode #148: Sort List
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None:
            return head
        slow, fast = head, head.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        l1, l2 = head, slow.next
        slow.next = None
        l1, l2 = self.sortList(l1), self.sortList(l2)
        dummy = ListNode()
        tail = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                tail.next = l1
                l1 = l1.next
            else:
                tail.next = l2
                l2 = l2.next
            tail = tail.next
        tail.next = l1 or l2
        return dummy.next

// Accepted solution for LeetCode #148: Sort List
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn sort_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        fn merge(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            match (l1, l2) {
                (None, Some(node)) | (Some(node), None) => Some(node),
                (Some(mut node1), Some(mut node2)) => {
                    if node1.val < node2.val {
                        node1.next = merge(node1.next.take(), Some(node2));
                        Some(node1)
                    } else {
                        node2.next = merge(Some(node1), node2.next.take());
                        Some(node2)
                    }
                }
                _ => None,
            }
        }

        fn sort(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            if head.is_none() || head.as_ref().unwrap().next.is_none() {
                return head;
            }
            let mut head = head;
            let mut length = 0;
            let mut cur = &head;
            while cur.is_some() {
                length += 1;
                cur = &cur.as_ref().unwrap().next;
            }
            let mut cur = &mut head;
            for _ in 0..length / 2 - 1 {
                cur = &mut cur.as_mut().unwrap().next;
            }
            let right = cur.as_mut().unwrap().next.take();

            merge(sort(head), sort(right))
        }
        sort(head)
    }
}

// Accepted solution for LeetCode #148: Sort List
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function sortList(head: ListNode | null): ListNode | null {
    if (head === null || head.next === null) {
        return head;
    }
    let [slow, fast] = [head, head.next];
    while (fast !== null && fast.next !== null) {
        slow = slow.next!;
        fast = fast.next.next;
    }
    let [l1, l2] = [head, slow.next];
    slow.next = null;
    l1 = sortList(l1);
    l2 = sortList(l2);
    const dummy = new ListNode();
    let tail = dummy;
    while (l1 !== null && l2 !== null) {
        if (l1.val <= l2.val) {
            tail.next = l1;
            l1 = l1.next;
        } else {
            tail.next = l2;
            l2 = l2.next;
        }
        tail = tail.next;
    }
    tail.next = l1 ?? l2;
    return dummy.next;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.