面试题17：合并两个排序的链表

注意某个链表为null的情况，合并链表会改变原链表结构

递归：

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);
    }
    else
    {
        pMergedHead = pHead2;
        pMergedHead->m_pNext = Merge(pHead1, pHead2->m_pNext);
    }

    return pMergedHead;
}

自己写的非递归：

先找到头结点，然后用head3遍历修改链表12每个节点的指向

ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
    if (pHead1 == NULL)
        return pHead2;
    if (pHead2 == NULL)
        return pHead1;
    ListNode *pMergeHead = NULL;
    if (pHead1->m_nValue < pHead2->m_nValue)
    {
        pMergeHead = pHead1;
        pHead1 = pHead1->m_pNext;
    }
    else
    {
        pMergeHead = pHead2;
        pHead2 = pHead2->m_pNext;
    }
    ListNode *pHead3 = pMergeHead;
    while (pHead1 != NULL && pHead2 != NULL)
    {
        if (pHead1->m_nValue < pHead2->m_nValue)
        {
            pHead3->m_pNext = pHead1;
            pHead3 = pHead3->m_pNext;
            pHead1 = pHead1->m_pNext;
        }
        else
        {
            pHead3->m_pNext = pHead2;
            pHead3 = pHead3->m_pNext;
            pHead2 = pHead2->m_pNext;
        }
    }
    if (pHead1 == NULL)
        pHead3->m_pNext = pHead2;
    else
        pHead3->m_pNext = pHead1;

    return pMergeHead;
}