剑指Offer第17题（合并两个排序的链表）

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题目描述：

       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。例如输入图3.7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：

struct ListNode
{
    int            m_nValue;
    ListNode*      m_pNext;
};

解法一：

 /*
  * @name   MergeList
  * @brief  合并两个升序链表为一个升序链表
  * @param  [in] ListNode * pHead1
  * @param  [in] ListNode * pHead2
  * @return ListNode*
  */
ListNode* MergeList(ListNode* pHead1, ListNode* pHead2)
{
    if (NULL == pHead1)
    {
        return pHead2;
    }
    if (NULL == pHead2)
    {
        return pHead1;
    }

    ListNode* pNode = NULL;
    ListNode* pNext = NULL;
    ListNode* pHead = NULL;
    while (pHead1 != NULL || pHead2 != NULL)
    {
        if (pHead1 == NULL)
        {
            pNext = pHead2;
            pHead2 = pHead2->m_pNext;
        }
        else if (pHead2 == NULL)
        {
            pNext = pHead1;
            pHead1 = pHead1->m_pNext;
        }
        else
        {
            if (pHead1->m_nValue <= pHead2->m_nValue)
            {
                pNext = pHead1;
                pHead1 = pHead1->m_pNext;
            }
            else
            {
                pNext = pHead2;
                pHead2 = pHead2->m_pNext;
            }
        }

        if (pNode == NULL)
        {
            pNode = pNext;   
            pHead = pNode;
        }
        else
        {
            pNode->m_pNext = pNext;
            pNode = pNext;
        }
    }
    return pHead;
}

解法二：递归实现

/*
* @name   MergeListRecursion
* @brief  合并两个升序链表为一个升序链表（递归实现）
* @param  [in] ListNode * pHead1
* @param  [in] ListNode * pHead2
* @return ListNode*
*/
ListNode* MergeListRecursion(ListNode* pHead1, ListNode* pHead2)
{
    if (NULL == pHead1)
    {
        return pHead2;
    }
    if (NULL == pHead2)
    {
        return pHead1;
    }

    ListNode* pHead = NULL;
    if (pHead1->m_nValue <= pHead2->m_nValue)
    {
        pHead = pHead1;
        pHead->m_pNext = MergeListRecursion(pHead1->m_pNext, pHead2);
    }
    else
    {
        pHead = pHead2;
        pHead->m_pNext = MergeListRecursion(pHead1, pHead2->m_pNext);
    }
    return pHead;
}

测试代码：

/*
 * @name   PrintList
 * @brief  打印链表
 * @param  [in] ListNode * pHead
 * @return void
 */
void PrintList(ListNode* pHead)
{
    if (NULL == pHead)
    {
        return;
    }
    while (pHead != NULL)
    {
        cout << pHead->m_nValue << " ";
        pHead = pHead->m_pNext;
    }
    cout << endl;
}


int _tmain(int argc, _TCHAR* argv[])
{
    //测试1题描述例子
    ListNode* pHead1 = new ListNode();
    pHead1->m_nValue = 1;
    ListNode* pTmp = new ListNode();
    pTmp->m_nValue = 3;
    pHead1->m_pNext = pTmp;

    pTmp->m_pNext = new ListNode();
    pTmp = pTmp->m_pNext;
    pTmp->m_nValue = 5;

    pTmp->m_pNext = new ListNode();
    pTmp = pTmp->m_pNext;
    pTmp->m_nValue = 7;

    ListNode* pHead2 = new ListNode();
    pHead2->m_nValue = 2;
    pTmp = new ListNode();
    pTmp->m_nValue = 4;
    pHead2->m_pNext = pTmp;

    pTmp->m_pNext = new ListNode();
    pTmp = pTmp->m_pNext;
    pTmp->m_nValue = 6;

    pTmp->m_pNext = new ListNode();
    pTmp = pTmp->m_pNext;
    pTmp->m_nValue = 8;

    PrintList(pHead1);
    PrintList(pHead2);
    cout << "测试题目第1个链表为空：" << endl;
    PrintList(MergeList(NULL, pHead2));
    cout << "测试题目第2个链表为空：" << endl;
    PrintList(MergeList(pHead1, NULL));
    cout << "测试题目2个链表为空：" << endl;
    PrintList(MergeList(NULL, NULL));
    cout << "测试题目2个链表都只有1个结点：" << endl;
    ListNode* pNode1 = new ListNode();
    pNode1->m_nValue = 9;
    ListNode* pNode2 = new ListNode();
    pNode2->m_nValue = 10;
    PrintList(MergeList(pNode1, pNode2));
    cout << "测试题目例子：" << endl;
    PrintList(MergeList(pHead1, pHead2));

    system("pause");
    return 0;
}

运行结果：

标签：ListNode,17,Offer,nValue,链表,pHead1,pHead2,pTmp,pNext
来源： https://blog.csdn.net/qq_23903863/article/details/103287076