剑指offer:合并两个排序的链表

题目描述

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

ListNode* pHead;
if(pHead1->val <= pHead2->val)
{
pHead = pHead1;
pHead1 = pHead1->next;
}
else
{
pHead = pHead2;
pHead2 = pHead2->next;
}
ListNode* pout =  pHead;

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

if(pHead1 == NULL)
{
while(pHead2 != NULL)
{
pHead->next = pHead2;
pHead2 = pHead2->next;
pHead =   pHead -> next;
}
}

if(pHead2 == NULL)
{
while(pHead1 != NULL)
{
pHead->next = pHead1;
pHead1 = pHead1->next;
pHead =   pHead -> next;
}
}
pHead -> next = NULL;
return pout;
}
};