--- id: 587d8251367417b2b2512c64 title: Search within a Linked List challengeType: 1 videoUrl: '' localeTitle: 在链接列表中搜索 --- ## Description
让我们为链表类添加一些更有用的方法。如果我们可以判断我们的列表是否为空,那么它是否有用,就像我们的StackQueue类一样?我们还应该能够在链表中找到特定元素。遍历数据结构是你想要进行大量练习的东西!让我们创建一个indexOf方法,该方法将element作为参数,并在链表中返回该元素的index 。如果在链接列表中找不到该元素,则返回-1 。让我们实现一个相反的方法:一个elementAt方法,它将index作为参数并返回给定index处的element 。如果未找到任何element ,则返回undefined
## Instructions
编写一个检查链表是否为空的isEmpty方法,返回给定元素indexindexOf方法,以及返回给定index.elementelementAt index.
## Tests
```yml tests: - text: 您的LinkedList类应该有一个indexOf方法。 testString: 'assert((function(){var test = new LinkedList(); return (typeof test.indexOf === "function")}()), "Your LinkedList class should have a indexOf method.");' - text: 您的LinkedList类应该有一个elementAt方法。 testString: 'assert((function(){var test = new LinkedList(); return (typeof test.elementAt === "function")}()), "Your LinkedList class should have a elementAt method.");' - text: 您的size方法应返回链表的长度 testString: 'assert((function(){var test = new LinkedList(); test.add("cat"); test.add("dog"); test.add("kitten"); return test.size() === 3}()), "Your size method should return the length of the linked list");' - text: indexOf方法应该返回给定元素的索引。 testString: 'assert((function(){var test = new LinkedList(); test.add("cat"); test.add("dog"); test.add("kitten"); return test.indexOf("kitten") === 2}()), "Your indexOf method should return the index of the given element.");' - text: 您的elementAt方法应该返回给定索引处的元素。 testString: 'assert((function(){var test = new LinkedList(); test.add("cat"); test.add("dog"); test.add("kitten"); return test.elementAt(1) === "dog"}()), "Your elementAt method should return at element at a given index.");' ```
## Challenge Seed
```js function LinkedList() { var length = 0; var head = null; var Node = function(element){ // {1} this.element = element; this.next = null; }; this.size = function() { return length; }; this.head = function(){ return head; }; this.add = function(element){ var node = new Node(element); if(head === null){ head = node; } else { currentNode = head; while(currentNode.next){ currentNode = currentNode.next; } currentNode.next = node; } length++; }; this.remove = function(element){ var currentNode = head; var previousNode; if(currentNode.element === element){ head = currentNode.next; } else { while(currentNode.element !== element) { previousNode = currentNode; currentNode = currentNode.next; } previousNode.next = currentNode.next; } length --; }; // Only change code below this line // Only change code above this line } ```
## Solution
```js // solution required ```