剑指offer之合并两个已排序好的链表

问题：输入两个单调递增的链表，输出两个链表合成后的链表，并满足合成后的链表单调不减规则。

思路：既然输入的链表已经单调递增，那就采取“递归”逐一比较合成是最直接的方法。

具体代码：（C++）

struct ListNode

{

    int val;

    ListNode* next;

    ListNode(int x):val(x),next(nullptr){}

};

class Solution

{

    public:

            ListNode*  Merge(ListNode* pHead1, LIstNode* pHead2)

            {

                    if(pHead1 == nullptr)

                            return pHead2;

                    if(pHead2 == nullptr)

                            return pHead1;

                    

                    ListNode*    pMerge = nullptr;

                    if(pHead1->val <= pHead2->val)

                    {

                            pMerge = pHead1;

                            pMerge->next = Merge(pHead1->next,pHead2);

                    }

                    else {

                            pMerge= pHead2;

                            pMerge->next = Merge(pHead1,pHead2->next);

                    }

                    return pMerge;

            }

};