707. Design Linked List

Problem Link

Description

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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.
• void addAtIndex(int index, int val) Add a node of value val before the indexth 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 indexth 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.

Solution

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Python3

class Node:
    def __init__(self, value):
        self.val = value
        self.next = None
 
class MyLinkedList:
 
    def __init__(self):
        self.head = None
        self.size = 0
 
    def get(self, index: int) -> int:
        if index < 0 or index >= self.size or not self.head: return -1
        
        curr = self.head
        for _ in range(index):
            curr = curr.next
        
        return curr.val
 
    def addAtHead(self, val: int) -> None:
        node = Node(val)
        node.next = self.head
        
        self.head = node
        self.size += 1
 
    def addAtTail(self, val: int) -> None:
        curr = self.head
        
        if not curr:
            self.head = Node(val)
        else:
            while curr.next:
                curr = curr.next
            curr.next = Node(val)
 
        self.size += 1
    
    def addAtIndex(self, index: int, val: int) -> None:
        if index == 0:
            self.addAtHead(val)
            return
        
        if index == self.size:
            self.addAtTail(val)
            return
        
        if not (0 <= index < self.size): return
        
        node = Node(val)
        curr = self.head
        
        for _ in range(index - 1):
            curr = curr.next
        
        nxt = curr.next
        curr.next = None
        node.next = nxt
        curr.next = node
        self.size += 1
        
 
    def deleteAtIndex(self, index: int) -> None:
        if index == 0:
            self.head = self.head.next
            self.size -= 1
            return
        
        if not (0 <= index < self.size): return
        
        curr = self.head
        
        for _ in range(index - 1):
            curr = curr.next
        
        curr.next = curr.next.next
        self.size -= 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)