劍指offer：合併兩個排序的鏈表

#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;
}