剑指offer--面试题17

题目：合并两个排序的单向链表

自己所写代码如下：

ListNode* MergeSortedLists(ListNode* pHead1, ListNode* pHead2)
{
    if(pHead1 == NULL || pHead2 == NULL)
        return NULL;
    return (pHead1->value < pHead2->value)? MergeTwoList(pHead1,pHead2):MergeTwoList(pHead2,pHead1);
}

ListNode* MergeTwoList(ListNode* pFirst,ListNode* pSecond)
{
    ListNode* pHead = pFirst;
    ListNode* ptemp = pSecond;

    while(pHead->pNext != NULL || ptemp->pNext != NULL)
    {
        while(pHead->pNext != NULL && pHead->pNext->value <= ptemp->value)
            pHead = pHead->pNext;

        ListNode* pHead1 = pHead->pNext;
        ListNode* ptemp1 = ptemp->pNext;
        ptemp->pNext = pHead->pNext;
        pHead->pNext = ptemp;
        ptemp = ptemp1;
        pHead = pHead1;
    }

    if(pHead->pNext == NULL)
        pHead->pNext = ptemp;

    return pFirst;
}

PS:未经测试！

作者分析的太精辟了！相比之下，自己的思路还不够完善，同时没有想到用递归实现。。。

//思路：两个链表中的头结点作比较，较小的被选为‘头结点’并链接到前一节点之后；
//      不断缩小规模，直到某个链表为空！
//所以递归！
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
    //递归终止条件
    //经典！
    if(pHead1 == NULL)
        return pHead2;
    else if(pHead2 == NULL)
        return pHead1;

    ListNode* pMerged = NULL;
    if(pHead1->value < pHead2->value)
    {
        pMerged = pHead1;
        pMerged->pNext = Merge(pHead1->pNext,pHead2);
    }
    else
    {
        pMerged = pHead2;
        pMerged->pNext = Merge(pHead1,pHead2->pNext);
    }

    return pMerged;

}