剑指offer--17.合并两个排序的链表

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

class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
//递归退出条件
if(pHead1==NULL)
return pHead2;
else if(pHead2==NULL)
return pHead1;

ListNode* pRes=NULL;

if(pHead1->val<=pHead2->val)
{
pRes=pHead1;
pRes->next=Merge(pHead1->next,pHead2);
}
else
{
pRes=pHead2;
pRes->next=Merge(pHead1,pHead2->next);
}
return pRes;
}
};

class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{

if(pHead1==NULL)
return pHead2;
else if(pHead2==NULL)
return pHead1;

//pRes返回链表的头结点
ListNode* pRes=NULL;
//pLast返回链表的尾节点
ListNode* pLast=NULL;

if(pHead1->val<=pHead2->val)
{
pRes=pHead1;
pHead1=pHead1->next;
}
else
{
pRes=pHead2;
pHead2=pHead2->next;
}
pLast=pRes;

while(pHead1!=NULL&&pHead2!=NULL)
{
if(pHead1->val<=pHead2->val)
{
pLast->next=pHead1;
pLast=pLast->next;
pHead1=pHead1->next;
}
else
{
pLast->next=pHead2;
pLast=pLast->next;
pHead2=pHead2->next;

}
}
if(pHead2==NULL)
pLast->next=pHead1;
else if(pHead1==NULL)
pLast->next=pHead2;

return pRes;
}
};