LeetCode #707 — MEDIUM

Design Linked List

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

The Problem

Problem Statement

Design your implementation of the linked list. You can choose to use a singly or doubly linked list.
A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node.
If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.

Implement the MyLinkedList class:

MyLinkedList() Initializes the MyLinkedList object.
int get(int index) Get the value of the index^th node in the linked list. If the index is invalid, return -1.
void addAtHead(int val) Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
void addAtTail(int val) Append a node of value val as the last element of the linked list.
void addAtIndex(int index, int val) Add a node of value val before the index^th node in the linked list. If index equals the length of the linked list, the node will be appended to the end of the linked list. If index is greater than the length, the node will not be inserted.
void deleteAtIndex(int index) Delete the index^th node in the linked list, if the index is valid.

Example 1:

Input
["MyLinkedList", "addAtHead", "addAtTail", "addAtIndex", "get", "deleteAtIndex", "get"]
[[], [1], [3], [1, 2], [1], [1], [1]]
Output
[null, null, null, null, 2, null, 3]

Explanation
MyLinkedList myLinkedList = new MyLinkedList();
myLinkedList.addAtHead(1);
myLinkedList.addAtTail(3);
myLinkedList.addAtIndex(1, 2);    // linked list becomes 1->2->3
myLinkedList.get(1);              // return 2
myLinkedList.deleteAtIndex(1);    // now the linked list is 1->3
myLinkedList.get(1);              // return 3

Constraints:

0 <= index, val <= 1000
Please do not use the built-in LinkedList library.
At most 2000 calls will be made to get, addAtHead, addAtTail, addAtIndex and deleteAtIndex.

Step 01

Brute Force Baseline

Problem summary: Design your implementation of the linked list. You can choose to use a singly or doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed. Implement the MyLinkedList class: MyLinkedList() Initializes the MyLinkedList object. int get(int index) Get the value of the indexth node in the linked list. If the index is invalid, return -1. void addAtHead(int val) Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. void addAtTail(int val) Append a node of value val as the last element of the linked list.

Baseline thinking

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

Pattern signal: Linked List · Design

Example 1

["MyLinkedList","addAtHead","addAtTail","addAtIndex","get","deleteAtIndex","get"]
[[],[1],[3],[1,2],[1],[1],[1]]

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 #707: Design Linked List
class MyLinkedList {
    private ListNode dummy = new ListNode();
    private int cnt;

    public MyLinkedList() {
    }

    public int get(int index) {
        if (index < 0 || index >= cnt) {
            return -1;
        }
        var cur = dummy.next;
        while (index-- > 0) {
            cur = cur.next;
        }
        return cur.val;
    }

    public void addAtHead(int val) {
        addAtIndex(0, val);
    }

    public void addAtTail(int val) {
        addAtIndex(cnt, val);
    }

    public void addAtIndex(int index, int val) {
        if (index > cnt) {
            return;
        }
        var pre = dummy;
        while (index-- > 0) {
            pre = pre.next;
        }
        pre.next = new ListNode(val, pre.next);
        ++cnt;
    }

    public void deleteAtIndex(int index) {
        if (index < 0 || index >= cnt) {
            return;
        }
        var pre = dummy;
        while (index-- > 0) {
            pre = pre.next;
        }
        var t = pre.next;
        pre.next = t.next;
        t.next = null;
        --cnt;
    }
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList obj = new MyLinkedList();
 * int param_1 = obj.get(index);
 * obj.addAtHead(val);
 * obj.addAtTail(val);
 * obj.addAtIndex(index,val);
 * obj.deleteAtIndex(index);
 */

// Accepted solution for LeetCode #707: Design Linked List
type MyLinkedList struct {
	dummy *ListNode
	cnt   int
}

func Constructor() MyLinkedList {
	return MyLinkedList{&ListNode{}, 0}
}

func (this *MyLinkedList) Get(index int) int {
	if index < 0 || index >= this.cnt {
		return -1
	}
	cur := this.dummy.Next
	for ; index > 0; index-- {
		cur = cur.Next
	}
	return cur.Val
}

func (this *MyLinkedList) AddAtHead(val int) {
	this.AddAtIndex(0, val)
}

func (this *MyLinkedList) AddAtTail(val int) {
	this.AddAtIndex(this.cnt, val)
}

func (this *MyLinkedList) AddAtIndex(index int, val int) {
	if index > this.cnt {
		return
	}
	pre := this.dummy
	for ; index > 0; index-- {
		pre = pre.Next
	}
	pre.Next = &ListNode{val, pre.Next}
	this.cnt++
}

func (this *MyLinkedList) DeleteAtIndex(index int) {
	if index < 0 || index >= this.cnt {
		return
	}
	pre := this.dummy
	for ; index > 0; index-- {
		pre = pre.Next
	}
	t := pre.Next
	pre.Next = t.Next
	t.Next = nil
	this.cnt--
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * obj := Constructor();
 * param_1 := obj.Get(index);
 * obj.AddAtHead(val);
 * obj.AddAtTail(val);
 * obj.AddAtIndex(index,val);
 * obj.DeleteAtIndex(index);
 */

# Accepted solution for LeetCode #707: Design Linked List
class MyLinkedList:
    def __init__(self):
        self.dummy = ListNode()
        self.cnt = 0

    def get(self, index: int) -> int:
        if index < 0 or index >= self.cnt:
            return -1
        cur = self.dummy.next
        for _ in range(index):
            cur = cur.next
        return cur.val

    def addAtHead(self, val: int) -> None:
        self.addAtIndex(0, val)

    def addAtTail(self, val: int) -> None:
        self.addAtIndex(self.cnt, val)

    def addAtIndex(self, index: int, val: int) -> None:
        if index > self.cnt:
            return
        pre = self.dummy
        for _ in range(index):
            pre = pre.next
        pre.next = ListNode(val, pre.next)
        self.cnt += 1

    def deleteAtIndex(self, index: int) -> None:
        if index >= self.cnt:
            return
        pre = self.dummy
        for _ in range(index):
            pre = pre.next
        t = pre.next
        pre.next = t.next
        t.next = None
        self.cnt -= 1


# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)

// Accepted solution for LeetCode #707: Design Linked List
#[derive(Default)]
struct MyLinkedList {
    head: Option<Box<ListNode>>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl MyLinkedList {
    fn new() -> Self {
        Default::default()
    }

    fn get(&self, mut index: i32) -> i32 {
        if self.head.is_none() {
            return -1;
        }
        let mut cur = self.head.as_ref().unwrap();
        while index > 0 {
            match cur.next {
                None => {
                    return -1;
                }
                Some(ref next) => {
                    cur = next;
                    index -= 1;
                }
            }
        }
        cur.val
    }

    fn add_at_head(&mut self, val: i32) {
        self.head = Some(Box::new(ListNode {
            val,
            next: self.head.take(),
        }));
    }

    fn add_at_tail(&mut self, val: i32) {
        let new_node = Some(Box::new(ListNode { val, next: None }));
        if self.head.is_none() {
            self.head = new_node;
            return;
        }
        let mut cur = self.head.as_mut().unwrap();
        while let Some(ref mut next) = cur.next {
            cur = next;
        }
        cur.next = new_node;
    }

    fn add_at_index(&mut self, mut index: i32, val: i32) {
        let mut dummy = Box::new(ListNode {
            val: 0,
            next: self.head.take(),
        });
        let mut cur = &mut dummy;
        while index > 0 {
            if cur.next.is_none() {
                return;
            }
            cur = cur.next.as_mut().unwrap();
            index -= 1;
        }
        cur.next = Some(Box::new(ListNode {
            val,
            next: cur.next.take(),
        }));
        self.head = dummy.next;
    }

    fn delete_at_index(&mut self, mut index: i32) {
        let mut dummy = Box::new(ListNode {
            val: 0,
            next: self.head.take(),
        });
        let mut cur = &mut dummy;
        while index > 0 {
            if let Some(ref mut next) = cur.next {
                cur = next;
            }
            index -= 1;
        }
        cur.next = cur.next.take().and_then(|n| n.next);
        self.head = dummy.next;
    }
}

// Accepted solution for LeetCode #707: Design Linked List
class LinkNode {
    public val: number;
    public next: LinkNode;

    constructor(val: number, next: LinkNode = null) {
        this.val = val;
        this.next = next;
    }
}

class MyLinkedList {
    public head: LinkNode;

    constructor() {
        this.head = null;
    }

    get(index: number): number {
        if (this.head == null) {
            return -1;
        }
        let cur = this.head;
        let idxCur = 0;
        while (idxCur < index) {
            if (cur.next == null) {
                return -1;
            }
            cur = cur.next;
            idxCur++;
        }
        return cur.val;
    }

    addAtHead(val: number): void {
        this.head = new LinkNode(val, this.head);
    }

    addAtTail(val: number): void {
        const newNode = new LinkNode(val);
        if (this.head == null) {
            this.head = newNode;
            return;
        }
        let cur = this.head;
        while (cur.next != null) {
            cur = cur.next;
        }
        cur.next = newNode;
    }

    addAtIndex(index: number, val: number): void {
        if (index <= 0) {
            return this.addAtHead(val);
        }
        const dummy = new LinkNode(0, this.head);
        let cur = dummy;
        let idxCur = 0;
        while (idxCur < index) {
            if (cur.next == null) {
                return;
            }
            cur = cur.next;
            idxCur++;
        }
        cur.next = new LinkNode(val, cur.next || null);
    }

    deleteAtIndex(index: number): void {
        if (index == 0) {
            this.head = (this.head || {}).next;
            return;
        }
        const dummy = new LinkNode(0, this.head);
        let cur = dummy;
        let idxCur = 0;
        while (idxCur < index) {
            if (cur.next == null) {
                return;
            }
            cur = cur.next;
            idxCur++;
        }
        cur.next = (cur.next || {}).next;
    }
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * var obj = new MyLinkedList()
 * var param_1 = obj.get(index)
 * obj.addAtHead(val)
 * obj.addAtTail(val)
 * obj.addAtIndex(index,val)
 * obj.deleteAtIndex(index)
 */

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.