更多剑指offer面试习题请点击：《剑指offer》（第二版）题集目录索引

1. 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。

2. 递归解法

ListNode* MergeR(ListNode* pHead1, ListNode* pHead2)
{
    if (pHead1 == nullptr)
        return pHead2;
    else if (pHead2 == nullptr)
        return pHead1;

    ListNode* pMergedHead = nullptr;

    if (pHead1->m_nValue < pHead2->m_nValue)
    {
        pMergedHead = pHead1;
        pMergedHead->m_pNext = MergeR(pHead1->m_pNext, pHead2);
    }
    else
    {
        pMergedHead = pHead2;
        pMergedHead->m_pNext = MergeR(pHead2->m_pNext, pHead1);
    }

    return pMergedHead;
}

3. 非递归解法

ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
    if (pHead1 == nullptr)
        return pHead2;
    else if (pHead2 == nullptr)
        return pHead1;

    ListNode* pMergedHead = nullptr;
    ListNode* pTail = nullptr;

    if (pHead1->m_nValue < pHead2->m_nValue)
    {
        pMergedHead = pHead1;
        pHead1 = pHead1->m_pNext;
    }
    else
    {
        pMergedHead = pHead2;
        pHead2 = pHead2->m_pNext;
    }
    pTail = pMergedHead;

    while (pHead1 != nullptr && pHead2 != nullptr)
    {
        if (pHead1->m_nValue < pHead2->m_nValue)
        {
            pTail->m_pNext = pHead1;
            pHead1 = pHead1->m_pNext;
        }
        else
        {
            pTail->m_pNext = pHead2;
            pHead2 = pHead2->m_pNext;
        }
        pTail = pTail->m_pNext;
    }
    if (pHead1 != nullptr)
    {
        pTail->m_pNext = pHead1;
    }
    else if (pHead2 != nullptr)
    {
        pTail->m_pNext = pHead2;
    }

    return pMergedHead;
}