浙大PAT 2-11. 两个有序链表序列的合并 （解题思路）

2-11. 两个有序链表序列的合并

1 3 5 -1
2 4 6 8 10 -1

1 2 3 4 5 6 8 10

1 2 3 4 5 -1
1 2 3 4 5 -1

1 1 2 2 3 3 4 4 5 5

-1
-1

NULL

#include <iostream>
#include <deque>
using namespace std;

deque<int> a, b, c;

int main()
{
int t;
scanf("%d", &t);
//	cin>>t;
while(t != -1)
{
a.push_back(t);
//		cin>>t;
scanf("%d", &t);
}
//	cin>>t;
scanf("%d", &t);
while(t != -1)
{
b.push_back(t);
scanf("%d", &t);
//		cin>>t;
}
if(a.size() + b.size() == 0)
{
cout<<"NULL"<<endl;
return 0;
}
while(a.size() && b.size())
{
if( a.front() < b.front())
{
c.push_back(a.front());
a.pop_front();
}
else
{
c.push_back(b.front());
b.pop_front();
}
}
while(a.size())
{
c.push_back(a.front());
a.pop_front();
}
while(b.size())
{
c.push_back(b.front());
b.pop_front();
}
bool first = true;
while(c.size())
{
if(first)
{
printf("%d", c.front());
//			cout<<c.front();
first = false;
}
else
printf(" %d", c.front());
//		cout<<" "<<c.front();
c.pop_front();
}
cout<<endl;
return 0;
}