合并两个排序的链表

时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M 

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

 

思路：

　　一、设立pPoint1指向第一个链表，pPoint2指向第二个链表，newPoint作为新链表的指针，比较pPoint1与pPoint2指针所指数据的大小，确定链表的next的指向

注：ListNode* pPoint1的指针的引用为ListNode*& pPoint1

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if(pHead1 == NULL)
            return pHead2;
        if(pHead2 == NULL)
            return pHead1;
        ListNode *pPoint1 = pHead1;
        ListNode *pPoint2 = pHead2;
        ListNode *newPoint ,*newPointTemp;
         
        newPoint = isMin(pPoint1,pPoint2);
       
        newPointTemp = newPoint;
         
        while(pPoint1 != NULL && pPoint2!= NULL)
        {
            newPointTemp->next = isMin(pPoint1,pPoint2);
            newPointTemp = newPointTemp->next;
        }
        if(pPoint1 == NULL )
            newPointTemp->next = pPoint2;
         else
             newPointTemp->next = pPoint1;
        return newPoint;
    }
     ListNode* isMin(ListNode*& pPoint1,ListNode*& pPoint2)
    {
         ListNode *temp;
         if(pPoint1->val <= pPoint2->val)
            {
                temp = pPoint1;
                pPoint1 = pPoint1->next;
                 
            }
            else
            {
                temp = pPoint2;
                pPoint2 = pPoint2->next;
            
            }
         return temp;
    }
};

二、采用递归的方式，较难理解

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