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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 | #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; } |
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