javascript数据结构与算法（一）单向链表与双向链表

摘要：创建双向链表创建节点继承添加前继指针初始化双向链表对链表最后一个元素进行引用插入操作尾插法任意位置添加元素删除操作删除特定位置的元素查询操作查询元素的位置查询尾部元素修改操作清空双向链表双向链表对象转为字符串

线性表 （List）：零个或者多个数据元素的有限序列。线性表是一个序列，具有一对一的关系，而不能具备一对多的关系。线性表要求有限性，数据类型也要相同。
本文参考的主要是大话数据结构的原理进行理解，用Javascript实现线性表的相关操作。

</>复制代码 
//创建节点
class Node{
    constructor(element){
        this.element = element;//数据域
        this.next = undefined;//指针域
    }
};
//判断传入的对象与链表的元素是否相等，待会实现indexOf()要用到，在这里先写上
function(a,b){
    return a == b;
}; 
//创建链表
class LinkedList {
  constructor(equalsFn = defaultEquals) {
    this.equalsFn = equalsFn;
    this.count = 0;
    this.head = undefined;
    }
};

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isEmpty() {
    return this.size() === 0;
  }

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size() {
    return this.count;
  }

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indexOf(element) {
    let current = this.head;
    for (let i = 0; i < this.size() && current != null; i++) {
      if (this.equalsFn(element, current.element)) {
        return i;
      }
      current = current.next;
    }
    return -1;
  }

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getHead() {
    return this.head;
  }

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getElementAt(index) {
    if (index >= 0 && index <= this.count) {
      let node = this.head;
      for (let i = 0; i < index && node != null; i++) {
        node = node.next;
      }
      return node;
    }
    return undefined;
  }

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push(element) {
    const node = new Node(element);
    let current;
    if (this.head == null) {
      // catches null && undefined
      this.head = node;
    } else {
      current = this.head;
      while (current.next != null) {
        current = current.next;
      }
      current.next = node;
    }
    this.count++;
  }

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insert(element, index) {
    if (index >= 0 && index <= this.count) {
      const node = new Node(element);
      if (index === 0) {
        const current = this.head;
        node.next = current;
        this.head = node;
      } else {
        const previous = this.getElementAt(index - 1);
        node.next = previous.next;
        previous.next = node;
      }
      this.count++;
      return true;
    }
    return false;
  }

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removeAt(index) {
    if (index >= 0 && index < this.count) {
      let current = this.head;
      if (index === 0) {
        this.head = current.next;
      } else {
        const previous = this.getElementAt(index - 1);
        current = previous.next;
        previous.next = current.next;
      }
      this.count--;
      return current.element;
    }
    return undefined;
  }

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remove(element) {
    const index = this.indexOf(element);
    return this.removeAt(index);
  }

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toString() {
    if (this.head == null) {
      return "";
    }
    let objString = `${this.head.element}`;
    let current = this.head.next;
    for (let i = 1; i < this.size() && current != null; i++) {
      objString = `${objString},${current.element}`;
      current = current.next;
    }
    return objString;
  }
}

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clear() {
    this.head = undefined;
    this.count = 0;
  }

以上就是单向链表的常见操作。

由于单向链表只能从头到尾的遍历，如果查询得是下一节点，单向表时间复杂度为O（1），但是要查询的是上一节点的话，那么时间复杂度就是O(n)。如果链表也可以正反遍历的话，那么查询操作的时间复杂度就都是O（1）啦，这时我们的前辈就提出了双向链表这一神奇的链表。由于双向链表是单向链表的拓展，只是多了一个指针，对于查询操作并没有帮助，所以实现方法还是跟单向链表一样，这里就不多加阐述。

</>复制代码 
//创建节点
class DoublyNode extends Node {
  constructor(element, next, prev) {
    super(element, next);//继承
    this.prev = prev;//添加前继指针
  }
};
//初始化双向链表
class DoublyLinkedList extends LinkedList {
  constructor(equalsFn = defaultEquals) {
    super(equalsFn);
    this.tail = undefined;//对链表最后一个元素进行引用
  }

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push(element) {
    const node = new DoublyNode(element);
    if (this.head == null) {
      this.head = node;
      this.tail = node; // NEW
    } else {
      // attach to the tail node // NEW
      this.tail.next = node;
      node.prev = this.tail;
      this.tail = node;
    }
    this.count++;
  }

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insert(element, index) {
    if (index >= 0 && index <= this.count) {
      const node = new DoublyNode(element);
      let current = this.head;
      if (index === 0) {
        if (this.head == null) { // NEW
          this.head = node;
          this.tail = node; // NEW
        } else {
          node.next = this.head;
          this.head.prev = node; // NEW
          this.head = node;
        }
      } else if (index === this.count) { // last item NEW
        current = this.tail;
        current.next = node;
        node.prev = current;
        this.tail = node;
      } else {
        const previous = this.getElementAt(index - 1);
        current = previous.next;
        node.next = current;
        previous.next = node;
        current.prev = node; // NEW
        node.prev = previous; // NEW
      }
      this.count++;
      return true;
    }
    return false;
  }

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removeAt(index) {
    if (index >= 0 && index < this.count) {
      let current = this.head;
      if (index === 0) {
        this.head = this.head.next;
        // if there is only one item, then we update tail as well //NEW
        if (this.count === 1) {
          // {2}
          this.tail = undefined;
        } else {
          this.head.prev = undefined;
        }
      } else if (index === this.count - 1) {
        // last item //NEW
        current = this.tail;
        this.tail = current.prev;
        this.tail.next = undefined;
      } else {
        current = this.getElementAt(index);
        const previous = current.prev;
        // link previous with current"s next - skip it to remove
        previous.next = current.next;
        current.next.prev = previous; // NEW
      }
      this.count--;
      return current.element;
    }
    return undefined;
  }

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indexOf(element) {
    let current = this.head;
    let index = 0;
    while (current != null) {
      if (this.equalsFn(element, current.element)) {
        return index;
      }
      index++;
      current = current.next;
    }
    return -1;
  }

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getTail() {
    return this.tail;
  }

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clear() {
    super.clear();
    this.tail = undefined;
  }

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inverseToString() {
    if (this.tail == null) {
      return "";
    }
    let objString = `${this.tail.element}`;
    let previous = this.tail.prev;
    while (previous != null) {
      objString = `${objString},${previous.element}`;
      previous = previous.prev;
    }
    return objString;
  }
}