双向链表

在计算机科学中, 一个 双向链表(doubly linked list) 是由一组称为节点的顺序链接记录组成的链接数据结构。每个节点包含两个字段，它们是对节点序列中上一个节点和下一个节点的引用。开始节点和结束节点的上一个链接和下一个链接分别指向某种终止节点，通常是前哨节点或null，以方便遍历列表。如果只有一个前哨节点，则列表通过前哨节点循环链接。它可以被概念化为两个由相同数据项组成的单链表，但顺序相反。

复杂度

时间复杂度

Access	Search	Insertion	Deletion
O(n)	O(n)	O(1)	O(1)

空间复杂度

O(n)

class DoublyLinkedListNode {
  constructor(value, next = null, previous = null) {
    this.value = value;
    this.next = next;
    this.previous = previous;
  }

  toString(callback) {
    return callback ? callback(this.value) : `${this.value}`;
  }
}

import DoublyLinkedListNode from "./DoublyLinkedListNode";
import Comparator from "../utils/Comparator";

export default class DoublyLinkedList {
  /**
   * @param {Function} [comparatorFunction]
   */
  constructor(comparatorFunction) {
    /** @var DoublyLinkedListNode */
    this.head = null;

    /** @var DoublyLinkedListNode */
    this.tail = null;

    this.compare = new Comparator(comparatorFunction);
  }

  /**
   * @param {*} value
   * @return {DoublyLinkedList}
   */
  prepend(value) {
    // Make new node to be a head.
    const newNode = new DoublyLinkedListNode(value, this.head);

    // If there is head, then it won't be head anymore.
    // Therefore, make its previous reference to be new node (new head).
    // Then mark the new node as head.
    if (this.head) {
      this.head.previous = newNode;
    }
    this.head = newNode;

    // If there is no tail yet let's make new node a tail.
    if (!this.tail) {
      this.tail = newNode;
    }

    return this;
  }

  /**
   * @param {*} value
   * @return {DoublyLinkedList}
   */
  append(value) {
    const newNode = new DoublyLinkedListNode(value);

    // If there is no head yet let's make new node a head.
    if (!this.head) {
      this.head = newNode;
      this.tail = newNode;

      return this;
    }

    // Attach new node to the end of linked list.
    this.tail.next = newNode;

    // Attach current tail to the new node's previous reference.
    newNode.previous = this.tail;

    // Set new node to be the tail of linked list.
    this.tail = newNode;

    return this;
  }

  /**
   * @param {*} value
   * @return {DoublyLinkedListNode}
   */
  delete(value) {
    if (!this.head) {
      return null;
    }

    let deletedNode = null;
    let currentNode = this.head;

    while (currentNode) {
      if (this.compare.equal(currentNode.value, value)) {
        deletedNode = currentNode;

        if (deletedNode === this.head) {
          // If HEAD is going to be deleted...

          // Set head to second node, which will become new head.
          this.head = deletedNode.next;

          // Set new head's previous to null.
          if (this.head) {
            this.head.previous = null;
          }

          // If all the nodes in list has same value that is passed as argument
          // then all nodes will get deleted, therefore tail needs to be updated.
          if (deletedNode === this.tail) {
            this.tail = null;
          }
        } else if (deletedNode === this.tail) {
          // If TAIL is going to be deleted...

          // Set tail to second last node, which will become new tail.
          this.tail = deletedNode.previous;
          this.tail.next = null;
        } else {
          // If MIDDLE node is going to be deleted...
          const previousNode = deletedNode.previous;
          const nextNode = deletedNode.next;

          previousNode.next = nextNode;
          nextNode.previous = previousNode;
        }
      }

      currentNode = currentNode.next;
    }

    return deletedNode;
  }

  /**
   * @param {Object} findParams
   * @param {*} findParams.value
   * @param {function} [findParams.callback]
   * @return {DoublyLinkedListNode}
   */
  find({ value = undefined, callback = undefined }) {
    if (!this.head) {
      return null;
    }

    let currentNode = this.head;

    while (currentNode) {
      // If callback is specified then try to find node by callback.
      if (callback && callback(currentNode.value)) {
        return currentNode;
      }

      // If value is specified then try to compare by value..
      if (value !== undefined && this.compare.equal(currentNode.value, value)) {
        return currentNode;
      }

      currentNode = currentNode.next;
    }

    return null;
  }

  /**
   * @return {DoublyLinkedListNode}
   */
  deleteTail() {
    if (!this.tail) {
      // No tail to delete.
      return null;
    }

    if (this.head === this.tail) {
      // There is only one node in linked list.
      const deletedTail = this.tail;
      this.head = null;
      this.tail = null;

      return deletedTail;
    }

    // If there are many nodes in linked list...
    const deletedTail = this.tail;

    this.tail = this.tail.previous;
    this.tail.next = null;

    return deletedTail;
  }

  /**
   * @return {DoublyLinkedListNode}
   */
  deleteHead() {
    if (!this.head) {
      return null;
    }

    const deletedHead = this.head;

    if (this.head.next) {
      this.head = this.head.next;
      this.head.previous = null;
    } else {
      this.head = null;
      this.tail = null;
    }

    return deletedHead;
  }

  /**
   * @return {DoublyLinkedListNode[]}
   */
  toArray() {
    const nodes = [];

    let currentNode = this.head;
    while (currentNode) {
      nodes.push(currentNode);
      currentNode = currentNode.next;
    }

    return nodes;
  }

  /**
   * @param {*[]} values - Array of values that need to be converted to linked list.
   * @return {DoublyLinkedList}
   */
  fromArray(values) {
    values.forEach(value => this.append(value));

    return this;
  }

  /**
   * @param {function} [callback]
   * @return {string}
   */
  toString(callback) {
    return this.toArray()
      .map(node => node.toString(callback))
      .toString();
  }

  /**
   * Reverse a linked list.
   * @returns {DoublyLinkedList}
   */
  reverse() {
    let currNode = this.head;
    let prevNode = null;
    let nextNode = null;

    while (currNode) {
      // Store next node.
      nextNode = currNode.next;
      prevNode = currNode.previous;

      // Change next node of the current node so it would link to previous node.
      currNode.next = prevNode;
      currNode.previous = nextNode;

      // Move prevNode and currNode nodes one step forward.
      prevNode = currNode;
      currNode = nextNode;
    }

    // Reset head and tail.
    this.tail = this.head;
    this.head = prevNode;

    return this;
  }
}