双向链表

复杂度

时间复杂度

获取：O(n)

查询：O(n)

插入：O(1)

删除：O(1)

空间复杂度

O(n)

DoublyLinkedListNode：

export default 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}`;
  }
}

DoublyLinkedList：

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) {//头插法1.头节点存在2.正常3.尾指针不存在
    // 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) {//尾插法，1.头节点不存在2.其他情况
    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)) {//有头必有尾，考虑尾部的情况
        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() {//转换成数组
    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) {//数组转链表
    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.
      //逆置。
      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;
  }
}

note:

1.

it('should prepend node to linked list', () => {
    const linkedList = new DoublyLinkedList();

    linkedList.prepend(2);
    expect(linkedList.head.toString()).toBe('2');
    expect(linkedList.tail.toString()).toBe('2');

    linkedList.append(1);
    linkedList.prepend(3);//3<- ->2<- ->1

    expect(linkedList.head.next.next.previous).toBe(linkedList.head.next);
    expect(linkedList.tail.previous.next).toBe(linkedList.tail);
    expect(linkedList.tail.previous.value).toBe(2);
    expect(linkedList.toString()).toBe('3,2,1');
  });

2.通过回调查找函数

it('should find node by callback', () => {
    const linkedList = new DoublyLinkedList();

    linkedList
      .append({ value: 1, key: 'test1' })
      .append({ value: 2, key: 'test2' })
      .append({ value: 3, key: 'test3' });

    const node = linkedList.find({ callback: value => value.key === 'test2' });

    expect(node).toBeDefined();
    expect(node.value.value).toBe(2);
    expect(node.value.key).toBe('test2');
    expect(linkedList.find({ callback: value => value.key === 'test5' })).toBeNull();
  });

3.双向链表逆置算法：

reverse() {
    let currNode = this.head;
    let prevNode = null;
    let nextNode = null;

    while (currNode) {
      // Store next node.找到前后结点。
      nextNode = currNode.next;
      prevNode = currNode.previous;

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

      // Move prevNode and currNode nodes one step forward.反复进行。
      prevNode = currNode;
      currNode = nextNode;
    }

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

    return this;
  }