学步园

将2个分别有序的链表合并成一个链表，并且合并后的链表依然有序。

方法：初始化一个头结点head，然后比较list1,和list2链表的第一个节点，选择比较小的连接到head上去，如此往复。

#include "stdafx.h"
#include<iostream>
#include<queue>
using namespace std;

typedef struct node{
        int data;
        struct node *next;
}Node,*List;

List createList(int N,int multi)
{
	List head = (List)malloc(sizeof(Node));
	head->data = 0;
	head->next=NULL;
	int count = 0;
	List p = head;
	while(count<N)
	{	count++;
		List s = (List)malloc(sizeof(Node));
		s->data = count*multi;
		s->next = NULL;
		p->next = s;
		p = s;
	}
	return head;
}

void traverse(List head)
{
	if(head == NULL)
	{
		return;
	}
	List p = head->next;
	while(p)
	{
		cout<<p->data<<" ";
		p = p->next;
	}
	cout<<endl;
}

List unionList(List list1,List list2)
{
	if(list1 == NULL||list1->next==NULL)
		return list2;
	if(list2 == NULL||list1->next==NULL)
		return list1;


	list1 = list1->next;
	list2 = list2->next;
	List head = (List)malloc(sizeof(Node));
	List p = NULL;
	
	if(list1!=NULL && list2!=NULL)//初始化第一个节点
	{
		if(list1->data<=list2->data)
		{
			head->next = list1;
			list1 = list1->next;
		}else{
			head->next = list2;
			list2 = list2->next;
		}
	}
	p = head->next;
	while(list1!=NULL && list2!=NULL)//主体部分
	{
		if(list1->data<=list2->data)
		{
			p->next = list1;
			list1 = list1->next;
			p = p->next;
		}else{
			p->next = list2;
			list2 = list2->next;
			p = p->next;
		}
	}
	while(list1 != NULL)//如果list2已经完了,list1还有剩余
	{
		p->next = list1;
		list1 = list1->next;
		p = p->next;
	}
	while(list2 != NULL)//如果list1已经完了,list2还有剩余
	{
		p->next = list2;
		list2 = list2->next;
		p = p->next;
	}
	return head;
}

int main()
{
	int N = 10;
	List head1 = createList(N,1);
	List head2 = createList(N,2);
	traverse(head1);
	traverse(head2);
	List head = unionList(head1,head2);
	traverse(head);	
    getchar();
    return 0;
}