剑指Offer面试题:16.合并两个排序的链表

PS:这也是一道出镜率极高的面试题,我相信很多童鞋都会很眼熟,就像于千万人之中遇见不期而遇的人,没有别的话可说,唯有轻轻地问一声:“哦,原来你也在这里? ”

一、题目:合并两个排序的链表

题目:输入两个递增排序的链表,合并这两个链表并使新链表中的结点仍然是按照递增排序的。例如输入下图中的链表1和链表2,则合并之后的升序链表如链表3所示。

  链表结点定义如下,使用C#描述:

    public class Node
    {
        public int Data { get; set; }
        // 指向后一个节点
        public Node Next { get; set; }

        public Node(int data)
        {
            this.Data = data;
        }

        public Node(int data, Node next)
        {
            this.Data = data;
            this.Next = next;
        }
    }

二、解题思路

  Step1.定义一个指向新链表的指针,暂且让它指向NULL;

  Step2.比较两个链表的头结点,让较小的头结点作为新链表的头结点;

  Step3.递归比较两个链表的其余节点,让较小的节点作为上一个新节点的后一个节点;

    public Node Merge(Node head1, Node head2)
    {
        if (head1 == null)
        {
            return head2;
        }
        else if (head2 == null)
        {
            return head1;
        }

        Node newHead = null;

        if (head1.Data <= head2.Data)
        {
            newHead = head1;
            newHead.Next = Merge(head1.Next, head2);
        }
        else
        {
            newHead = head2;
            newHead.Next = Merge(head1, head2.Next);
        }

        return newHead;
    }

三、单元测试

3.1 测试准备

  (1)借助MSUnit框架进行初始化与清理工作[TestInitialize]与[TestCleanup]

    private MergeHelper mergeHelper;

    [TestInitialize]
    public void Initialize()
    {
        // 实例化
        mergeHelper = new MergeHelper();
    }

    [TestCleanup]
    public void CleanUp()
    {
        // 不用TA了
        mergeHelper = null;
    } 

  (2)封装一个便于测试对比的辅助方法,将新链表生成一个字符串用于对比

    public string GetListString(Node head)
    {
        if (head == null)
        {
            return null;
        }

        StringBuilder sbList = new StringBuilder();
        while (head != null)
        {
            sbList.Append(head.Data.ToString());
            head = head.Next;
        }

        return sbList.ToString();
    } 

3.2 测试用例

  (1)功能测试

    // list1: 1->3->5
    // list2: 2->4->6
    [TestMethod]
    public void MergeTest1()
    {
        Node node1 = new Node(1);
        Node node3 = new Node(3);
        Node node5 = new Node(5);

        node1.Next = node3;
        node3.Next = node5;

        Node node2 = new Node(2);
        Node node4 = new Node(4);
        Node node6 = new Node(6);

        node2.Next = node4;
        node4.Next = node6;

        Node newHead = mergeHelper.Merge(node1, node2);
        Assert.AreEqual(GetListString(newHead), "123456");
    }

    // 两个链表中有重复的数字
    // list1: 1->3->5
    // list2: 1->3->5
    [TestMethod]
    public void MergeTest2()
    {
        Node node1 = new Node(1);
        Node node3 = new Node(3);
        Node node5 = new Node(5);

        node1.Next = node3;
        node3.Next = node5;

        Node node2 = new Node(1);
        Node node4 = new Node(3);
        Node node6 = new Node(5);

        node2.Next = node4;
        node4.Next = node6;

        Node newHead = mergeHelper.Merge(node1, node2);
        Assert.AreEqual(GetListString(newHead), "113355");
    } 

  (2)特殊输入测试

    // 两个链表都只有一个数字
    // list1: 1
    // list2: 2
    [TestMethod]
    public void MergeTest3()
    {
        Node node1 = new Node(1);
        Node node2 = new Node(2);

        Node newHead = mergeHelper.Merge(node1, node2);
        Assert.AreEqual(GetListString(newHead), "12");
    }

    // 两个链表中有一个空链表
    // list1: 1->3->5
    // list2: null
    [TestMethod]
    public void MergeTest4()
    {
        Node node1 = new Node(1);
        Node node3 = new Node(3);
        Node node5 = new Node(5);

        node1.Next = node3;
        node3.Next = node5;

        Node newHead = mergeHelper.Merge(node1, null);
        Assert.AreEqual(GetListString(newHead), "135");
    }

    // 两个链表都是空链表
    // list1: null
    // list2: null
    [TestMethod]
    public void MergeTest5()
    {
        Node newHead = mergeHelper.Merge(null, null);
        Assert.AreEqual(GetListString(newHead), null);
    } 

3.3 测试结果

  (1)测试通过情况

  (2)代码覆盖率

 

posted @ 2015-08-30 20:09 Edison Chou 阅读(...) 评论(...) 编辑 收藏