题目描述
blablabla
样例
#include <iostream>
using namespace std;
const int N = 1e5+10;
int e[N], ne[N], head, idex;
void init(){
head = -1;
idex = 0;
}
void inhead(int x){
e[idex] = x;
ne[idex] = head;
head = idex++;
}
void dele(int k){
ne[k] = ne[ne[k]];
}
void insert(int k, int x){
e[idex] = x;
ne[idex] = ne[k];
ne[k] = idex++;
}
int main(){
int m;
scanf("%d", &m);
init();
char c;
while(m--){
//scanf("%c", &c);
cin >> c;
if(c == 'H'){
int x;
scanf("%d", &x);
inhead(x);
}
if(c == 'D'){
int k;
scanf("%d", &k);
if(k == 0) head = ne[head];
else dele(k-1);
}
if(c == 'I'){
int k, x;
scanf("%d%d", &k, &x);
insert(k-1, x);
}
}
for(int i = head; i != -1; i = ne[i]) printf("%d ", &e[i]);
return 0;
}
算法1
(暴力枚举) $O(n^2)$
blablabla
时间复杂度
参考文献
C++ 代码
blablabla
算法2
(暴力枚举) $O(n^2)$
blablabla
时间复杂度
参考文献
C++ 代码
blablabla