从整数数组中创建单链表

I'm having trouble with taking an array of integers and creating a singly linked list with JavaScript. It sounds easy enough, but there's something I'm just not seeing with the function I have an I'd appreciate any help you can provide.

我在使用整数数组并使用JavaScript创建单链表时遇到了麻烦。这听起来很容易，但有些东西，我只是没有看到我的功能，我很感激你能提供的任何帮助。

This is the constructor function I use to make nodes:

这是我用来制作节点的构造函数：

 function ListNode(val) {
     this.val = val;
     this.next = null;
 }

And this is the function I'm writing that is supposed to take an array and create a linked list out of it. The basic idea is just a while loop that shifts off the first value until there's nothing left to shift:

这是我正在编写的函数，它应该采用一个数组并从中创建一个链表。基本的想法只是一个while循环，它会移开第一个值，直到没有任何东西可以移动：

var createLinkedList = function(array) {
    var head = new ListNode(parseInt(array[0]));
    array.shift();
    while(array.length) {
        var prev = new ListNode(parseInt(array[0]));
        head.next = head;
        prev = head;
        array.shift();
    }
    return head;
}

I've tried running this with a couple basic arrays and it always just returns the last value in the array instead of a linked list. Is there something simple I'm just not seeing here? Thanks in advance.

我尝试使用几个基本数组运行它，它总是只返回数组中的最后一个值而不是链表。有什么简单的我只是没有看到这里？提前致谢。

1 个解决方案

#1

The problem is not with array.shift, but how you link the nodes together inside the while loop.

问题不在于array.shift，而是在while循环中如何将节点链接在一起。

To chain the nodes together, essentially you need to do:

要将节点链接在一起，基本上你需要做：

var new_node = new ListNode(parseInt(array[0]));
head.next = new_node;
new_node = new ListNode(parseInt(array[1]));
head.next.next = new_node;
new_node = new ListNode(parseInt(array[2]));
head.next.next.next = new_node;
.
.
.

I think you get the idea. So what you want to do is rework your while loop, store a reference of the tailNode so instead of calling head.next.next.next.next.next = newNode, you can call tailNode.next = newNode, and tailNode = newNode inside the while loop.

我想你应该已经明白了。所以你要做的是重做你的while循环，存储tailNode的引用，而不是调用head.next.next.next.next.next = newNode，你可以在里面调用tailNode.next = newNode和tailNode = newNode while循环。

#1