Problem:
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

Solution in JAVA (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	static Node nil = new Node();

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		Main main = new Main();
		main.init();
		for (int i = 0; i < n; i++) {
			String command = sc.next();
			int key = -1;
			if (sc.hasNextInt()) {
				key = sc.nextInt();
			}
			switch (command) {
			case "insert":
				main.insert(key);
				break;
			case "delete":
				main.deleteKey(key);
				break;
			case "deleteFirst":
				main.deleteFirst();
				break;
			case "deleteLast":
				main.deleteLast();
				break;
			}
		}
		// リストのすべての要素を表示
		Node cur = nil.next;
		StringBuilder sb = new StringBuilder();
		for (int i = 0;; i++) {
			if (cur == nil)
				break;
			if (i == 0) {
				sb.append(cur.key);
			} else {
				sb.append(" " + cur.key);
			}
			cur = cur.next;
		}
		System.out.println(sb);

	}

	// リストの初期化
	void init() {
		nil.next = nil;
		nil.prev = nil;

	}

	// 要素の挿入
	void insert(int key) {
		Node x = new Node(key);
		x.next = nil.next;
		nil.next.prev = x;
		x.prev = nil;
		nil.next = x;

	}

	// 要素の探索
	Node listSearch(int key) {
		Node cur = nil.next;
		while (cur!= nil && cur.key != key) { // keyが一致するまで
			cur = cur.next;
		}
		return cur;
	}

	// 特定のノードを削除
	void deleteNode(Node t) {
		if (t != nil) { // 削除するノードが番兵でないこと
			t.prev.next = t.next;
			t.next.prev = t.prev;
			t = null;
		}
	}

	// 先頭の要素を削除
	void deleteFirst() {
		deleteNode(nil.next);
	}

	// 末尾の要素を削除
	void deleteLast() {
		deleteNode(nil.prev);
	}

	// 特定のキーを持つ要素を削除
	void deleteKey(int key) {
		// 検索したノードを削除
		deleteNode(listSearch(key));
	}

}

class Node {
	int key;
	Node prev;
	Node next;

	public Node() {
	}

	public Node(int key) {
		this.key = key;
	}

}