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
code
/*
^....0
^ .1 ^1^
.. 01
1.^ 1.0
^ 1 ^ ^0.1
1 ^ ^..^
0. ^ 0^
.0 1 .^
.1 ^0 .........001^
.1 1. .111100....01^
00 ^ 11^ ^1. .1^
1.^ ^0 0^
.^ ^0..1
.1 1..^
1 .0 ^ ^
^ 00. ^^0.^
1 ^ 0 ^^110.^
0. 0 ^ ^^^10.01
^^ 010^ 1 1 ^^^1110.1
0001 10 0 ^ 1.1 ^^^1111110
0^ 10 . 01 ^^ ^^ ^^^1111^1.^ ^^^
10 10^ 0^ ^^111^^^0.1^ 1....^
11 0 ^^11^^^ 0.. ....1^ ^ ^
1. 0^ ^11^^^ ^ 1 111^ ^ 0.
10 00 11 ^^^^^ 1 0 1.
0^ ^0 ^0 ^^^^ 0 0.
0^ 1.0 .^ ^^^^ 1 1 .0
^.^ ^^ 0^ ^1 ^^^^ 0. ^.1
1 ^ 11 1. ^^^ ^ ^ ..^
^..^ ^1 ^.^ ^^^ .0 ^.0
0..^ ^0 01 ^^^ .. 0..^
1 .. .1 ^.^ ^^^ 1 ^ ^0001
^ 1. 00 0. ^^^ ^.0 ^.1
. 0^. ^.^ ^.^ ^^^ ..0.0
1 .^^. .^ 1001 ^^ ^^^ . 1^
. ^ ^. 11 0. 1 ^ ^^ 0.
0 ^. 0 ^0 1 ^^^ 0.
0.^ 1. 0^ 0 .1 ^^^ ..
.1 1. 00 . .1 ^^^ ..
1 1. ^. 0 .^ ^^ ..
0. 1. .^ . 0 .
.1 1. 01 . . ^ 0
^.^ 00 ^0 1. ^ 1 1
.0 00 . ^^^^^^ .
.^ 00 01 ..
1. 00 10 1 ^
^.1 00 ^. ^^^ .1
.. 00 .1 1..01 ..
1.1 00 1. ..^ 10
^ 1^ 00 ^.1 0 1 1
.1 00 00 ^ 1 ^
. 00 ^.^ 10^ ^^
1.1 00 00 10^
..^ 1. ^. 1.
0 1 ^. 00 00 .^
^ ^. ^ 1 00 ^0000^ ^ 01
1 0 ^. 00.0^ ^00000 1.00.1 11
. 1 0 1^^0.01 ^^^ 01
.^ ^ 1 1^^ ^.^
1 1 0.
.. 1 ^
1 1
^ ^ .0
1 ^ 1
.. 1.1 ^0.0
^ 0 1..01^^100000..0^
1 1 ^ 1 ^^1111^ ^^
0 ^ ^ 1 1000^
.1 ^.^ . 00
.. 1.1 0. 0
1. . 1. .^
1. 1 1. ^0
^ . ^.1 00 01
^.0 001. .^
*/
// Virtual_Judge —— Doubly Linked List Aizu - ALDS1_3_C.cpp created by VB_KoKing on 2019-04-29:12.
/* Procedural objectives：
Variables required by the program:
Procedural thinking：
Functions required by the program:
*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first hurricane that passed through your ear is you,
The first star you see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/
#include <cstdio>
#include <string>
#include <cstdlib>
#include <iostream>
using namespace std;
struct Node {
int key;
Node *next, *prev;
};
Node *nil;
void init() {
nil = (Node *) malloc(sizeof(Node));
nil->next = nil;
nil->prev = nil;
}
Node *list_search(int key) {
Node *cur = nil->next;
while (cur != nil && cur->key != key)
cur = cur->next;
return cur;
}
void print_list() {
Node *cur = nil->next;
int isf = 0;
while (true) {
if (cur == nil) break;
if (isf++ > 0) cout << ' ';
printf("%d", cur->key);
cur = cur->next;
}
cout << endl;
}
void delete_Node(Node *t) {
if (t == nil) return;
t->prev->next = t->next;
t->next->prev = t->prev;
free(t);
}
void delete_first() { delete_Node(nil->next); }
void delete_last() { delete_Node(nil->prev); }
void delete_key(int key) { delete_Node(list_search(key)); }
void insert(int key) {
Node *x = (Node *) malloc(sizeof(Node));
x->key = key;
x->next = nil->next;
nil->next->prev = x;
nil->next = x;
x->prev = nil;
}
int main() {
init();
string com;
int key, n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
cin >> com;
switch (com[0]) {
case 'i':
cin >> key;
insert(key);
break;
case 'd':
if (com.length() > 6) {
switch (com[6]) {
case 'F':
delete_first();
break;
case 'L':
delete_last();
break;
}
} else {
cin >> key;
delete_key(key);
}
break;
}
}
print_list();
return 0;
}