剑指offer：合并两个排序的链表

#include <iostream>
#include <stdio.h>
#include <malloc.h>

using namespace std;

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 * res = NULL;
ListNode * tmp = NULL;
while(pHead1 && pHead2)
{
if(pHead1->val > pHead2->val)
{
if(res == NULL)
res = tmp = pHead2;
else
{
res->next = pHead2;
res = res->next;
}
pHead2 = pHead2->next;
}
else
{
if(res == NULL)
res = tmp = pHead1;
else
{
res->next = pHead1;
res = res->next;
}
pHead1 = pHead1->next;
}
}
if(pHead1 == NULL)
res->next = pHead2;
if(pHead2 == NULL)
res->next = pHead1;
return tmp;
}
};

int main()
{
int i;
ListNode * p = (ListNode*)malloc(sizeof(ListNode));
p->val = 1;
p->next = NULL;
ListNode * p1 = p;
ListNode * q = (ListNode*)malloc(sizeof(ListNode));
q->val = 1;
q->next = NULL;
ListNode * q1 = q;
for(i = 2; i < 5; i++)
{
p->next = (ListNode*)malloc(sizeof(ListNode));
p = p->next;
p->val = i;
p->next = NULL;
q->next = (ListNode*)malloc(sizeof(ListNode));
q = q->next;
q->val = i;
q->next = NULL;
}
Solution s;
ListNode * node = s.Merge(p1, q1);
while(node)
{
cout << node->val << " ";
node = node->next;
}
return 0;
}