合并两个排序的链表

      问题: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。链表结点定义如下:

   struct  ListNode

   {

         int  m_nValue;

         ListNode *m_nNext;

    }

   

详细的代码实现如下:

//合并两个升序的链表
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
    if(pHead1 == NULL)
        return pHead2;
    else if(pHead2 == NULL)
        return pHead1;

    ListNode* pMergedHead = NULL;

    if(pHead1->m_nValue < pHead2->m_nValue)
    {
        pMergedHead = pHead1;
        pMergedHead->m_pNext = Merge(pHead1->m_pNext, pHead2);//递归调用
    }
    else
    {
        pMergedHead = pHead2;
        pMergedHead->m_pNext = Merge(pHead1, pHead2->m_pNext);//递归调用
    }

    return pMergedHead;
}

// ====================测试代码====================
ListNode* Test(char* testName, ListNode* pHead1, ListNode* pHead2)
{
    if(testName != NULL)
        printf("%s begins:\n", testName);

    printf("The first list is:\n");
    PrintList(pHead1);

    printf("The second list is:\n");
    PrintList(pHead2);

    printf("The merged list is:\n");
    ListNode* pMergedHead = Merge(pHead1, pHead2);
    PrintList(pMergedHead);
    
    printf("\n\n");

    return pMergedHead;
}

// list1: 1->3->5
// list2: 2->4->6
void Test1()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode3);
    ConnectListNodes(pNode3, pNode5);

    ListNode* pNode2 = CreateListNode(2);
    ListNode* pNode4 = CreateListNode(4);
    ListNode* pNode6 = CreateListNode(6);

    ConnectListNodes(pNode2, pNode4);
    ConnectListNodes(pNode4, pNode6);

    ListNode* pMergedHead = Test("Test1", pNode1, pNode2);

    DestroyList(pMergedHead);
}

// 两个链表中有重复的数字
// list1: 1->3->5
// list2: 1->3->5
void Test2()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode3);
    ConnectListNodes(pNode3, pNode5);

    ListNode* pNode2 = CreateListNode(1);
    ListNode* pNode4 = CreateListNode(3);
    ListNode* pNode6 = CreateListNode(5);

    ConnectListNodes(pNode2, pNode4);
    ConnectListNodes(pNode4, pNode6);

    ListNode* pMergedHead = Test("Test2", pNode1, pNode2);

    DestroyList(pMergedHead);
}

// 两个链表都只有一个数字
// list1: 1
// list2: 2
void Test3()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode2 = CreateListNode(2);

    ListNode* pMergedHead = Test("Test3", pNode1, pNode2);

    DestroyList(pMergedHead);
}

// 一个链表为空链表
// list1: 1->3->5
// list2: 空链表
void Test4()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode3);
    ConnectListNodes(pNode3, pNode5);

    ListNode* pMergedHead = Test("Test4", pNode1, NULL);

    DestroyList(pMergedHead);
}

// 两个链表都为空链表
// list1: 空链表
// list2: 空链表
void Test5()
{
    ListNode* pMergedHead = Test("Test5", NULL, NULL);
}

int _tmain(int argc, _TCHAR* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();

    return 0;
}

运行结果如下: