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。

// test14.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<iostream>
#include<fstream>
#include<string>
#include<cctype>
#include <vector>
#include<exception>
#include <initializer_list>
#include<stack>
#include <algorithm>

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)
	{
		ListNode *newHead = pHead1, *temp;
		vector<int>vec;
		if (pHead1 == NULL)
			return pHead2;
		if (pHead2 == NULL)
			return pHead1;
		
		while (pHead1->next!=NULL)
		{
			pHead1 = pHead1->next;
		}

		pHead1->next = pHead2;
		temp = newHead;

		while (newHead!=NULL)
		{
			vec.push_back(newHead->val);
			newHead = newHead->next;
		}
		
		newHead = temp;
		sort(vec.begin(),vec.end());

		for (auto it = vec.begin(); it != vec.end(); it++)
		{
			temp->val = *it;
			temp = temp->next;
		}
		
		return newHead;

	}
};

int main()
{
	vector<int> vec;
	Solution so;

	
	ListNode first(1);
	ListNode second(4);
	ListNode third(5);
	ListNode four(6);
	ListNode *head=&first;
	first.next = &second;
	second.next = &third;
	third.next = &four;


	ListNode first1(1);
	ListNode second1(2);
	ListNode third1(3);
	ListNode four1(7);
	ListNode *head1 = &first1;
	first1.next = &second1;
	second1.next = &third1;
	third1.next = &four1;



	
	ListNode *result = so.Merge(head, head1);
	while (result !=NULL)
	{
		cout << (*result).val<<"  ";
		result = (*result).next;
	}

	cout << endl;
	
	return 0;
}