剑指offer面试题-合并两个有序链表

• 题目：
  输入两个递增有序的链表，合并这两个链表并使新链表中的结点仍然是有序的，具体示例如下图，链表结点定义如下：

struct ListNoe{    ListNoe()    :_next(NULL)    , _data(0)    {}    ListNoe *_next;    int _data;};

解法一：循环

ListNoe * Merge(ListNoe *pHead1, ListNoe *pHead2){    if (!pHead1)        return pHead2;    if (!pHead2)        return pHead1;    //两个链表一定不为空    ListNoe *pHead3 = NULL;    ListNoe *pPre = NULL;    if (pHead1->_data <= pHead2->_data)    {        pHead3 = pHead1;        pPre = pHead1;        pHead1 = pHead1->_next;    }    else if (pHead1->_data > pHead2->_data)    {        pHead3 = pHead2;        pPre = pHead2;        pHead2 = pHead2->_next;    }    while (pHead1 && pHead2)    {        if (pHead1->_data <= pHead2->_data)        {            pPre->_next = pHead1;            pPre = pHead1;            pHead1 = pHead1->_next;        }        else if (pHead1->_data > pHead2->_data)        {            pPre->_next = pHead2;            pPre = pHead2;            pHead2 = pHead2->_next;        }    }    if (!pHead1)    {        pPre->_next = pHead2;    }    if (!pHead2)        pPre->_next = pHead1;    return pHead3;}

解法二：递归

ListNoe * Merge(ListNoe *pHead1, ListNoe *pHead2){    if (!pHead1)        return pHead2;    if (!pHead2)        return pHead1;    ListNoe *pPre = NULL;    if (pHead1->_data <= pHead2->_data)    {        pPre = pHead1;        pPre->_next = Merge(pHead1->_next, pHead2);    }    else    {        pPre = pHead2;        pPre->_next = Merge(pHead1, pHead2->_next);    }    return pPre;}

完整代码详见
https://coding.net/u/Hyacinth_Dy/p/MyCode/git/blob/master/%E5%90%88%E5%B9%B6%E6%9C%89%E5%BA%8F%E9%93%BE%E8%A1%A8%E5%92%8Co1%E4%B8%8B%E6%B1%821%E5%8A%A0%E5%88%B0n%E7%9A%84%E5%92%8C7-13