Leetcode #707: Design Linked List
In this guide, we solve Leetcode #707 Design Linked List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design your implementation of the linked list. You can choose to use a singly or doubly linked list.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Linked List
Intuition
Linked list problems often require pointer manipulation rather than extra memory.
Two-pointer techniques expose cycles, midpoints, or reordering patterns.
Approach
Traverse with fast/slow pointers or reverse sublists when needed.
Maintain invariants carefully to avoid losing nodes.
Steps:
- Traverse with pointers.
- Reverse or split if required.
- Reconnect nodes correctly.
Example
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
Python Solution
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)
Complexity
The time complexity is O(n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.