Achieve a single linked list is initially empty, supports three operations:
(1) a number to the list head insert;
(2) deleting the k-th number of digits after insertion;
(3) inserting a number in the k-th number of inserted
Now, after M times to operate the chain, complete all operations, the output of the entire list from beginning to end.
Note : k-th title number does not refer to the insertion of the k number of the current list. During operation, for example, a total of n number is inserted, the insert according to the chronological order, the order of the n number of: inserting a first number, the second number is inserted, ... n-th number of insertion.
Input Format
The first row contains an integer M, it represents the number of operations.
Next M rows, each row comprising an operation command, an operation command may be as following:
(1) "H x", x denotes a number from the list is inserted into the head.
(2) "D k", represents the number of deleted behind the k-th input number (when k is 0, means to delete the first node).
(3) "I kx", inserting a number x indicates the number of the input after the k-th (k in this operation are greater than 0).
Output Format
Total row, the entire chain from beginning to end output.
data range
1 ≤ M ≤ 100000 1≤M≤100000
all operations to ensure legal.
Sample input:
10
H 9
I 1 1
D 1
D 0
H 6
I 3 6
I 4 5
I 4 5
I 3 4
D 6
Sample output:
6 4 6 5
#include<iostream>
using namespace std;
const int N = 1e5 + 50;
int head,e[N],ne[N],idx;
void init(){
head = -1;
idx = 0;
}
void head_add(int x){
e[idx] = x;
ne[idx] = head;
head = idx++;
}
void d_k(int k){
ne[k] = ne[ne[k]];
}
void add_k(int k,int x){
e[idx] = x;
ne[idx] = ne[k];
ne[k] = idx++;
}
int main(){
int m,k;
int n;
scanf("%d",&m);
char op[2];
init();
while(m --){
scanf("%s",op);
if(op[0] == 'H'){
scanf("%d",&n);
head_add(n);
}else if(op[0] == 'D'){
scanf("%d",&k);
if(!k) head = ne[head];
else d_k(k - 1);
}else {
scanf("%d%d",&k,&n);
add_k(k - 1,n);
}
}
for(int i = head;i != -1;i = ne[i]) printf("%d ",e[i]);
return 0;
}