面试遇到的题目及解决办法

1.用js写一个类似Dom的getElementByTagName的函数。

试验了一下，js的childNodes属性有兼容性和准确度的问题，经常会多一些空白标签，所以用深搜的方法根本行不通，没法建树，没法准确获取节点树信息。

所以，想到了正则表达式，只用几行代码即可搞定。

            window.onload = function () {
            var reg = new RegExp("<span.+/span>", "g");
            var element = new Array();
            var temp;
            while ((temp = reg.exec(document.body.innerHTML)) != null) {
                element.push(temp);
            }
            while ((temp = element.pop()) != null) {
                alert(temp);
            }
        }    

之后，又想到另一种DFS深搜解法

     class Program
    {
        private static XmlElement _root;
        public static XmlElement Root
        {
            get { return _root; }
            set { _root = value; }
        }

        private static List<XmlNode> list = new List<XmlNode>();
        static void Main(string[] args)
        {
            WebClient wc = new WebClient();
            string htmlStr = wc.DownloadString("http://localhost:30907/test1.html");
            XmlDocument doc = new XmlDocument();
            doc.LoadXml(htmlStr);
            Root = doc.DocumentElement;
            
            DFS(Root);
            foreach (var n in list)
            {
                Console.WriteLine(n.OuterXml);
            }
            
            Console.ReadKey();
        }

        private static void DFS(XmlNode parent)
        {
            
            if (!parent.HasChildNodes)
            {
                return;
            }
            else if (parent.HasChildNodes)
            {
                for (int j = 0; j < parent.ChildNodes.Count; j++)
                {
                    if (parent.ChildNodes[j].Name == "span")
                    {
                        list.Add(parent.ChildNodes[j]);
                    }
                    DFS(parent.ChildNodes[j]);
                }
            }
        }
    } 

如果用宽搜BFS也可以，时间复杂度都是o(n)

    class Program
    {
        private static List<XmlNode> list = new List<XmlNode>();
        private static Queue<XmlNode> queue = new Queue<XmlNode>();
        static void Main(string[] args)
        {
            WebClient wc = new WebClient();
            string htmlStr = wc.DownloadString("http://localhost:30907/test1.html");
            XmlDocument doc = new XmlDocument();
            doc.LoadXml(htmlStr);
            XmlElement root = doc.DocumentElement;
            BFS(root);

            foreach (var n in list)
            {
                Console.WriteLine(n.OuterXml);
            }

            Console.ReadKey();
        }

        public static void BFS(XmlNode parent)
        {
            queue.Enqueue(parent);
            while (queue.Count != 0)
            {
                XmlNode n = queue.Dequeue();
                if (n.Name == "span")
                {
                    list.Add(n);
                }
                for (int i = 0; i < n.ChildNodes.Count; i++)
                {
                    queue.Enqueue(n.ChildNodes[i]);
                }
            }
        }
    }

2.用js写一个链表，或者用其他语言。

我比较熟C#所以就用C#写，类似List。

    public class Node
    {
        public int Num{get;set;}
        public Node Next{get;set;}
    }

    public class MyList
    {
        public Node Head { get; set; }
        public int Length { get; set; }

        public MyList()
        {
            Head = new Node();
            Head.Next = null;
            Length = 1;
        }

        public void Add(int num)
        {
            Node n = new Node();
            n.Num = num;
            Node node = Head;
            while (node.Next != null)
            {
                node = node.Next;
            }
            node.Next = n;
            Length++;
        }

        public void Delete(int index)
        {
            Node n = Head;
            if (index == 0)
            {
                Head = n.Next;
            }
            else if(Length-1==index)
            {
                for (int i = 0; i < index - 1; i++)
                {
                    n = n.Next;
                }
                n.Next = null;
            }
            else
            {
                for (int i = 0; i < index - 1; i++)
                {
                    n = n.Next;
                }
                n.Next = n.Next.Next;
            }
            Length--;
        }

        public int this[int index]
        {
            get {
                Node n = Head;
                for (int i = 0; i < index; i++)
                {
                    n = n.Next;
                }
                return n.Num;
            }
            set  {
                Node n = Head;
                for (int i = 0; i < index; i++)
                {
                    n = n.Next;
                }
                n.Num = value;
            }
        }
    }