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JavaScript Data Structure and Algorithm Linked List_Basic Knowledge

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Release: 2016-05-16 15:16:54
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Introduction to linked lists

Linked list is a common data structure and is also a linear list, but it does not store data in linear order. Instead, in each node, the pointer to the next node is stored. You can understand by looking at the picture. (Those with a C language background may have an easier time understanding it).
The use of a linked list structure can overcome the shortcoming of an array that requires the data size to be known in advance (an array in C language requires a predefined length). The linked list structure can make full use of computer memory space and achieve flexible dynamic memory management.
Next, we will introduce two common linked lists: one-way linked list and the implementation of doubly linked list in JavaScript.

One-way linked list

The simplest form of a linked list is a one-way linked list. The nodes in the linked list contain two parts. The first part stores its own information, and the second part stores a pointer to the next node. The last node points to NULL:

Implementation of one-way linked list in JavaScipt

First, create a constructor.

/**
 * 单向链表构造函数
 */
function LinkedList() {
 /**
  * 单向链表中节点的构造函数
  * @param {Any} element 要传入链表的节点
  */
 var Node = function(element) {
  this.element = element;
  //下个节点的地址
  this.next = null;
 }

 //单向链表的长度
 var length = 0;
 //单向链表的头结点,初始化为NULL
 var head = null;
}

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It is not difficult to see that the one-way linked list constructor is much more complicated than stacks and queues.
A one-way linked list requires the following methods:

  1. append(element): Add element to the end of the linked list
  2. insert(position,element): Insert an element into a certain position in the one-way linked list
  3. indexOf(element): Find the position of an element in a one-way linked list
  4. remove(element): Remove the given element
  5. removeAt(position): Remove an element at a certain position in the one-way linked list
  6. getHead(): Get the head of the one-way linked list
  7. isAmpty(): Checks whether the one-way linked list is empty, returns true if it is empty
  8. toString(): Output all the contents of the linked list as a string
  9. size(): Returns the length of the one-way linked list

append method:

Description: Add elements to the end of the one-way linked list.
Implementation:

/**
 * 向单向链表尾部添加元素
 * @param {Any} element 要加入链表的节点
 */
this.append = function(element) {
 var node = new Node(element);
 var current;

 if (head == null) {
  head = node;
 } else {
  // 当前项等于链表头部元素.
  // while循环到最后一个,从而将该节点加入链表尾部。
  current = head;
  // 当next为null时,判定为false。退出循环。
  while (current.next) {
   current = current.next;
  }
  current.next = node;
 }
 length++;
};

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insert method:

Description: Insert an element into a certain position in the one-way linked list.
Implementation:

/**
 * 向单向链表中插入某个元素
 * @param {Number} position 要插入的位置
 * @param {Any} element 要插入的元素
 * @return {Boolean}     插入成功返回true,失败返回false
 */
this.insert = function(position, element) {
 if (position >= 0 && position <= length) {
  var node = new Node(element);
  var current = head;
  var previous;
  var index = 0;

  if (position == 0) {
   node.next = current;
   head = node;
  } else {
   while (index++ < position) {
    previous = current;
    current = current.next;
   }

   previous.next = node;
   node.next = current;
  }

  length++;
  return true;
 } else {
  return false;
 }
};

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indexOf method:

Description: Find the position of an element in a one-way linked list.
Implementation:

/**
 * 寻找某个元素在单向链表中的位置
 * @param {Any} element 要寻找的元素
 * @return {Number}     返回值>=0则代表找到相应位置
 */
this.indexOf = function(element) {
 var current = head;
 var index = -1;

 while (current) {
  if (element === current.element) {
   return index;
  }
  index++;
  current = current.next;
 }

 return -1;
};

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remove method:

Description: Remove the given element.
Implementation:

/**
 * 移除给定的元素
 * @param {Any} element 要移除的元素
 * @return {Number}     返回值>=0表示移除成功
 */
this.remove = function(element) {
 var index = this.indexOf(element);
 return this.removeAt(index);
};
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removeAt method:

Description: Remove an element at a certain position in the one-way linked list.
Implementation:

/**
 * 移除单向链表中某一个元素
 * @param {Number} position 要移除元素的位置
 * @return {Any}     移除成功返回被移除的元素,不成功则返回NULL
 */
this.removeAt = function(position) {
 if (position > -1 && position < length) {
  var current = head;
  var previous;
  var index = 0;

  if (position == 0) {
   // 因为之前head指向第一个元素,现在把head修改为指向第二个元素。
   // 核心概念在于链表前后全靠指针链接,而非数组一般。
   // 所以只需要改变head的元素。
   head = current.next;
  } else {
   while (index++ < position) {
    // previous指要操作元素位置之前的那个元素,current表示之后的那个元素。
    previous = current;
    current = current.next;
   }

   previous.next = current.next;
  }

  length--;

  return current.element;
 } else {
  return null;
 }
};

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getHead method:

Description: Get the head of the one-way linked list.
Implementation:

/**
 * 获取单向链表的头部
 * @return {Any} 单向链表的头部
 */
this.getHead = function() {
 return head;
}
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isAmpty, toString, size methods

Implementation:

/**
 * 判断单向链表是否为空
 * @return {Boolean} 为空则返回true,不为空则返回false
 */
this.isAmpty = function() {
 return length === 0
};

/**
 * 将链表所有内容以字符串输出
 * @return {String} 要输出的字符串
 */
this.toString = function() {
 var current = head;
 var string = '';

 while (current) {
  string += current.element;
  current = current.next;
 }
 return string;
};

/**
 * 返回单向链表长度
 * @return {Number} 单向链表的长度
 */
this.size = function() {
 return length;
};

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Double linked list

Doubly linked lists are very similar to one-way linked lists. In a one-way linked list, there is only a link to the next node. But in a doubly linked list, there is a link to the previous node, which is bidirectional.

Implementation of doubly linked list in JavaScipt

First, it’s still the constructor:

/**
 * 双向链表的构造函数
 */
function DoublyLinkedList() {
 /**
  * 双向链表中节点的构造函数
  * @param {Any} element 要传入链表的元素
  */
 var Node = function(element) {
  this.element = element;
  this.prev = null;
  this.next = null;
 }

 //双向链表的长度
 var length = 0;
 //双向链表的头结点,初始化为NULL
 var head = null;
 //双向链表的尾结点,初始化为NULL
 var tail = null;
}

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Doubly linked lists need the following methods:

  1. append(element): Add elements to the end of the doubly linked list
  2. insert(position,element): Insert an element into a certain position in the doubly linked list
  3. removeAt(position): Remove an element at a certain position in the doubly linked list
  4. showHead(): Get the head of the doubly linked list
  5. showLength(): Get the length of the doubly linked list
  6. showTail(): Get the tail of the doubly linked list

append method:

Description: Add elements to the end of the doubly linked list
Implementation:

/**
 * 向链表尾部添加元素
 * @param {Any} element 要加入链表的节点
 * @return {Any}     加入链表的节点
 */
this.append = function(element) {
 var node = new Node(element);

 if (head === null) {
  head = node;
  tail = node;
 } else {
  var previous;
  var current = head;

  while (current.next) {
   current = current.next;
  }

  current.next = node;
  node.prev = current;
  tail = node;
 }

 length++;
 return node;
};

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insert method:

Description: Insert an element into a certain position in the doubly linked list.
Implementation:

/**
 * 向链表中插入某个元素
 * @param {Number} position 要插入的位置
 * @return {Boolean}     插入成功返回true,失败返回false
 */
this.insert = function(position, element) {
 if (position >= 0 && position <= length) {

  var node = new Node(element);
  var index = 0;
  var previous;
  var current = head;

  if (position === 0) {

   if (head === null) {
    head = node;
    tail = node;
   } else {
    current.prev = node;
    node.next = current;
    head = node;
   }
  } else if (position === length) {

   current = tail;
   current.next = node;
   node.prev = current;
   tail = node;
  } else {

   while (index++ < position) {
    previous = current;
    current = current.next;
   }

   previous.next = node;
   node.prev = previous;
   current.prev = node;
   node.next = current;
  }

  length++;
  return true;
 } else {
  return false;
 }
};

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removeAt method:

Description: Remove an element at a certain position in the doubly linked list.
Implementation:

/**
 * 移除链表中某一个元素
 * @param {Number} position 要移除元素的位置
 * @return {Any}       移除成功返回被移除的元素,不成功则返回false
 */
this.removeAt = function(position) {
 if (position > -1 && position < length) {
  var current = head;
  var index = 0;
  var previous;

  if (position === 0) {
   head = current.next;

   if (length === 1) {
    tail = null;
    head.prev = null;
   }
  } else if (position === length - 1) {
   current = tail;
   tail = current.prev;
   tail.next = null;
  } else {
   while (index++ < position) {
    previous = current.prev;
    current = current.next;
   }
   previous.next = current.next;
   current.next.prev = previous;
  }

  length--;
  return current.element;
 } else {
  return false;
 }
};

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showHead, showLength, showTail methods

Implementation:

/**
 * 获取链表的头部
 * @return {Any} 链表的头部
 */
this.showHead = function() {
 return head;
};

/**
 * 获取链表长度
 * @return {Number} 链表长度
 */
this.showLength = function() {
 return length;
};

/**
 * 获取链表尾部
 * @return {Any} 链表尾部
 */
this.showTail = function() {
 return tail;
};

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Impressions

In this section on linked lists, basically all code is written according to requirements first, and then compared with the book after writing. It was found that he was reduced to rubbish in an instant. There are many hidden pits in my own writing, and the logic is also very confusing. It seems that he is still too young.

source:php.cn
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