Doubly Linked List

本文蒐集

• LeetCode 707

但注意，雖然是同一題，但之前是實做 Singly Linked List ,本篇則是實做 Doubly Linked List

LeetCode 707: Design Linked List

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.

我的答案

class Node(object):

    def __init__(self, val):
        self.val = val
        self.next = None
        self.perv = None


class MyLinkedList:
    
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.head = None
        self.tail = None
        self.size = 0
        

    def get(self, index: int) -> int:
        """
        Get the value of the index-th node in the linked list. If the index is invalid, return -1.
        """
        if index < 0 or index >= self.size:
            return -1
            
        curr = self.head
        for i in range(index):
            curr = curr.next
        return curr.val
    
        
    def addAtHead(self, val: int) -> None:
        """
        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.
        """
        node = Node(val)
        node.next = self.head
        node.prev = None
        if self.size > 0:
            self.head.perv = node
        self.head = node
        self.size += 1
            
            
    def addAtTail(self, val: int) -> None:
        """
        Append a node of value val to the last element of the linked list.
        """
        if self.head is None:
            self.addAtHead(val)
            return
        
        node = Node(val)
        curr = self.head
        while curr.next:
            curr = curr.next
        node.perv = curr
        node.next = None
        curr.next = node
        self.size += 1
        

    def addAtIndex(self, index: int, val: int) -> None:
        """
        Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
        """
        if index < 0 or index > self.size:
            return
        
        if index == self.size:
            self.addAtTail(val)
            return
        
        if index == 0:
            self.addAtHead(val)
            return
        
        node = Node(val)
        curr = self.head
        for i in range(index-1):
            curr = curr.next
        node.next = curr.next
        node.perv = curr
        curr.next.perv = node
        curr.next = node
        self.size += 1
        

    def deleteAtIndex(self, index: int) -> None:
        """
        Delete the index-th node in the linked list, if the index is valid.
        """
        if self.size == 0:
            return
        
        if index < 0 or index >= self.size:
            return
        
        if index == 0:
            self.head = self.head.next
            if self.head:
                self.head.perv = None
            return
        
        curr = self.head
        for i in range(index-1):
            curr = curr.next
        curr.next = curr.next.next
        if index != self.size-1:
            curr.next.perv = curr
        self.size -= 1