数组实现单双链表的快速操作[时间复杂度为O(1)]

826. 单链表 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/828/

#include <iostream>
using namespace std;

const int N = 1e5+10;
int e[N],ne[N],head,idx;

int main()
{
    int n,k,x;
    char op;
    
    head = -1, idx = 0;
    cin >> n;
    while (n -- )
    {
        cin >> op;
        if(op == 'H'){
            cin >> x;
            e[idx] = x;
            ne[idx] = head;
            head = idx;
            idx ++;
        }else if(op == 'D'){
            cin >> k;
            if(!k) head = ne[head];
            else ne[k-1] = ne[ne[k-1]];
        }else if(op == 'I'){
            cin >> k >> x;
            e[idx] = x;
            ne[idx] = ne[k-1];
            ne[k-1] = idx;
            idx ++;
        }
    }
    
    for(int i = head;i != -1; i=ne[i]) cout << e[i] << ' ';
    return 0;
}

827. 双链表 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/description/829/

#include <iostream>
using namespace std;

const int N = 1e5+10;
int l[N],r[N],e[N],head,idx;

void insert(int k, int x)
{
    e[idx] = x;
    l[idx] = k, r[idx] = r[k];
    l[r[idx]] = idx, r[k] = idx++;
}

void del(int k)
{
    l[r[k]] = l[k];
    r[l[k]] = r[k];
}
int main()
{
    int n,k,x;
    cin >> n;
    string op;
    
    l[1] = 0,r[0] = 1;
    idx = 2;
    while (n -- )
    {
        cin >> op;
        if(op == "L"){
            cin >> x;
            insert(0, x);
        }
        else if(op == "R"){
            cin >> x;
            insert(l[1], x);
        }
        else if(op == "D"){
            cin >> k;
            del(k+1);
        }
        else if(op == "IL"){
            cin >> k >> x;
            insert(l[k+1], x);
        }
        else if(op == "IR"){
            cin >> k >> x;
            insert(k+1, x);
        }
    }
    
    for (int i = r[0]; i != 1; i=r[i]) cout << e[i] << ' ';
    return 0;
}