freeCodeCamp/curriculum/challenges/chinese/08-coding-interview-prep/data-structures/add-elements-at-a-specific-...

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id title challengeType videoUrl localeTitle
587d8252367417b2b2512c67 Add Elements at a Specific Index in a Linked List 1 在链接列表中的特定索引处添加元素

Description

让我们创建一个addAtindexelement方法在给定的索引处添加一个元素。就像我们如何删除给定索引处的元素一样我们需要在遍历链表时跟踪currentIndex。当currentIndex与给定索引匹配时我们需要重新分配上一个节点的下一个属性以引用新添加的节点。并且新节点应该引用currentIndex中的下一个节点。回到康加线的例子一个新人想加入这条线但他想加入中间。你处于中间位置所以你要把手从你前面的人身上移开。新人走过去把手放在你曾经牵过手的那个人身上现在你已经掌握了新人。说明创建addAtindexelement方法该方法在给定索引处添加元素。如果无法添加元素则返回false。注意请记住检查给定索引是否为负数或者是否长于链接列表的长度。

Instructions

Tests

tests:
  - text: 当给定索引为0时 <code>addAt</code>方法应重新分配<code>head</code>到新节点。
    testString: 'assert((function(){var test = new LinkedList(); test.add("cat"); test.add("dog"); test.addAt(0,"cat"); return test.head().element === "cat"}()), "Your <code>addAt</code> method should reassign <code>head</code> to the new node when the given index is 0.");'
  - text: 对于添加到链接列表的每个新节点, <code>addAt</code>方法应该将链表的长度增加一。
    testString: 'assert((function(){var test = new LinkedList(); test.add("cat"); test.add("dog"); test.addAt(0,"cat"); return test.size() === 3}()), "Your <code>addAt</code> method should increase the length of the linked list by one for each new node added to the linked list.");'
  - text: 如果无法添加节点,则<code>addAt</code>方法应返回<code>false</code> 。
    testString: 'assert((function(){var test = new LinkedList(); test.add("cat"); test.add("dog"); return (test.addAt(4,"cat") === false); }()), "Your <code>addAt</code> method should return <code>false</code> if a node was unable to be added.");'

Challenge Seed

function LinkedList() {
  var length = 0;
  var head = null;

  var Node = function(element){
    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++;
  };

  // Only change code below this line

  // Only change code above this line

}

Solution

// solution required