LeetCode #147 — MEDIUM

Insertion Sort List

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

The Problem

Problem Statement

Given the head of a singly linked list, sort the list using insertion sort, and return the sorted list's head.

The steps of the insertion sort algorithm:

Insertion sort iterates, consuming one input element each repetition and growing a sorted output list.
At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list and inserts it there.
It repeats until no input elements remain.

The following is a graphical example of the insertion sort algorithm. The partially sorted list (black) initially contains only the first element in the list. One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration.

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]

Constraints:

The number of nodes in the list is in the range [1, 5000].
-5000 <= Node.val <= 5000

Step 01

Brute Force Baseline

Problem summary: Given the head of a singly linked list, sort the list using insertion sort, and return the sorted list's head. The steps of the insertion sort algorithm: Insertion sort iterates, consuming one input element each repetition and growing a sorted output list. At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list and inserts it there. It repeats until no input elements remain. The following is a graphical example of the insertion sort algorithm. The partially sorted list (black) initially contains only the first element in the list. One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration.

Baseline thinking

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

Pattern signal: Linked List

Example 1

[4,2,1,3]

Example 2

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

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 #147: Insertion 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 insertionSortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(head.val, head);
        ListNode pre = dummy, cur = head;
        while (cur != null) {
            if (pre.val <= cur.val) {
                pre = cur;
                cur = cur.next;
                continue;
            }
            ListNode p = dummy;
            while (p.next.val <= cur.val) {
                p = p.next;
            }
            ListNode t = cur.next;
            cur.next = p.next;
            p.next = cur;
            pre.next = t;
            cur = t;
        }
        return dummy.next;
    }
}

// Accepted solution for LeetCode #147: Insertion Sort List
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func insertionSortList(head *ListNode) *ListNode {
	if head == nil || head.Next == nil {
		return head
	}
	dummy := &ListNode{head.Val, head}
	pre, cur := dummy, head
	for cur != nil {
		if pre.Val <= cur.Val {
			pre = cur
			cur = cur.Next
			continue
		}
		p := dummy
		for p.Next.Val <= cur.Val {
			p = p.Next
		}
		t := cur.Next
		cur.Next = p.Next
		p.Next = cur
		pre.Next = t
		cur = t
	}
	return dummy.Next
}

# Accepted solution for LeetCode #147: Insertion 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 insertionSortList(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        dummy = ListNode(head.val, head)
        pre, cur = dummy, head
        while cur:
            if pre.val <= cur.val:
                pre, cur = cur, cur.next
                continue
            p = dummy
            while p.next.val <= cur.val:
                p = p.next
            t = cur.next
            cur.next = p.next
            p.next = cur
            pre.next = t
            cur = t
        return dummy.next

// Accepted solution for LeetCode #147: Insertion Sort List
struct Solution;
use rustgym_util::*;

trait Insertion {
    fn insert(self, link: ListLink) -> ListLink;
}

impl Insertion for ListLink {
    fn insert(self, mut link: ListLink) -> ListLink {
        let val = link.as_ref().unwrap().val;
        if let Some(mut node) = self {
            if node.val > val {
                link.as_mut().unwrap().next = Some(node);
                link
            } else {
                node.next = node.next.take().insert(link);
                Some(node)
            }
        } else {
            link
        }
    }
}

impl Solution {
    fn insertion_sort_list(mut head: ListLink) -> ListLink {
        let mut prev = None;
        while let Some(mut node) = head {
            head = node.next.take();
            prev = prev.insert(Some(node));
        }
        prev
    }
}

#[test]
fn test() {
    let head = list!(4, 2, 1, 3);
    let res = list!(1, 2, 3, 4);
    assert_eq!(Solution::insertion_sort_list(head), res);
    let head = list!(-1, 5, 3, 4, 0);
    let res = list!(-1, 0, 3, 4, 5);
    assert_eq!(Solution::insertion_sort_list(head), res);
}

// Accepted solution for LeetCode #147: Insertion Sort List
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #147: Insertion 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 insertionSortList(ListNode head) {
//         if (head == null || head.next == null) {
//             return head;
//         }
//         ListNode dummy = new ListNode(head.val, head);
//         ListNode pre = dummy, cur = head;
//         while (cur != null) {
//             if (pre.val <= cur.val) {
//                 pre = cur;
//                 cur = cur.next;
//                 continue;
//             }
//             ListNode p = dummy;
//             while (p.next.val <= cur.val) {
//                 p = p.next;
//             }
//             ListNode t = cur.next;
//             cur.next = p.next;
//             p.next = cur;
//             pre.next = t;
//             cur = t;
//         }
//         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)

Space

O(1)

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.