剑指offer(8)-合并两个有序链表

题目：
输入两个递增排序的有序链表，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，链表结点的定义如下：

struct ListNode {    int val;    struct ListNode *next;    ListNode(int x) :            val(x), next(NULL) {    }};

简单地用递归实现，主要是注意写出递归出口条件，细心处理。

class Solution {public:    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)    {        if (pHead1 == NULL)            return pHead2;        if (pHead2 == NULL)            return pHead1;        ListNode *pMergedList = NULL;        if (pHead1->val <= pHead2->val)            {            pMergedList = pHead1;            pMergedList->next = Merge(pHead1->next,pHead2);        }        else            {            pMergedList = pHead2;            pMergedList->next = Merge(pHead1,pHead2->next);        }        return pMergedList;}};