AcWing 826. 单链表
原题链接
简单
作者:
pingsheng
,
2021-01-11 22:17:31
,
所有人可见
,
阅读 357
/*
head -> 9(1) -> 1(2) -> oo
head -> 9(1) -> oo
head -> oo
head -> 6(3)
head -> 6(3) -> 6(4) -> oo
head -> 6(3) -> 6(4) -> 5(5) -> oo
head -> 6(3) -> 6(4) -> 5(6) -> 5(5) -> oo
head -> 6(3) -> 4(7) -> 6(4) -> 5(6) -> 5(5) -> oo
head -> 6(3) -> 4(7) -> 6(4) -> 5(5) -> oo
*/
#include <iostream>
using namespace std;
const int N = 100010;
int head, e[N], ne[N], idx;
void init()
{
head = -1;
idx = 0;
}
void add_to_head(int x)
{
e[idx] = x, ne[idx] = head, head = idx ++;
}
void add(int k, int x)
{
e[idx] = x, ne[idx] = ne[k], ne[k] = idx ++;
}
void remove(int k)
{
ne[k] = ne[ne[k]];
}
int main()
{
int m;
cin >> m;
init();
while(m -- )
{
int k, x;
char op;
cin >> op;
if(op == 'H')
{
cin >> x;
add_to_head(x);
}
else if (op == 'D')
{
cin >> k;
if (!k) head = ne[head];
else remove(k - 1);
}
else
{
cin >> k >> x;
add(k - 1, x);
}
}
for(int i = head; i != -1; i = ne[i]) cout << e[i] << ' ';
cout << endl;
return 0;
}