Doubly Linked List

Doubly Linked List

Your task is to implement a double linked list.

Write a program which performs the following operations:

• insert x: insert an element with key x into the front of the list.
• delete x: delete the first element which has the key of x from the list. If there is not such element, you need not do anything.
• deleteFirst: delete the first element from the list.
• deleteLast: delete the last element from the list.

Input

The input is given in the following format:

n
command1 
command2 
...
commandn 

In the first line, the number of operations n is given. In the following n lines, the above mentioned operations are given in the following format:

• insert x
• delete x
• deleteFirst
• deleteLast

Output

Print all the element (key) in the list after the given operations. Two consequtive keys should be separated by a single space.

Constraints

• The number of operations ≤ 2,000,000
• The number of delete operations ≤ 20
• 0 ≤ value of a key ≤ 109
• The number of elements in the list does not exceed 106
• For a delete, deleteFirst or deleteLast operation, there is at least one element in the list.

Sample Input 1

7
insert 5
insert 2
insert 3
insert 1
delete 3
insert 6
delete 5

Sample Output 1

6 1 2

Sample Input 2

9
insert 5
insert 2
insert 3
insert 1
delete 3
insert 6
delete 5
deleteFirst
deleteLast

Sample Output 2

1

注意: 指令和数字不能同时输入, 比如deleteFirst只有指令没有数字, 数字要在相应的判断条件中输入

#include <iostream>
#include <list>
#include <cstdio>
using namespace std;

int main()
{
	int n, num;
	list<int> l;
	char s[20];
	scanf("%d", &n);
	for(int i = 0; i < n; ++ i)
	{
		scanf("%s", s);
		if(s[0] == 'i')
		{
			scanf("%d", &num);
			l.push_front(num);
		}
		else if(s[6] == 'F')
		{
			l.pop_front();
		}
		else if(s[6] == 'L')
		{
			l.pop_back();
		}
		else if(s[0] == 'd')
		{
			scanf("%d", &num);
			for(list<int>::iterator it = l.begin(); it != l.end(); it ++)
			{
				if(*it == num)
				{
					l.erase(it);
					break;
				}
			}
		}
	}
	
	int i = 0;
	for(list<int>::iterator it = l.begin(); it != l.end(); it ++)
	{
		if(i ++)	printf(" ");
		printf("%d", *it);
	}
	printf("\n");
	
	
	return 0;
}