数组模拟单链表/双链表

调试漫谈：如果用vscode调试一定要保证路径和文件名没有中文

单链表

这个比较简单，注意插入在某个数的右边，应该先将“待改动”的数作为被赋值数。也就是说一定要先ne[idx] = head / ne[k],再考虑ne[k]/head = idx

init: null -> -1 

after: x -> x -> x ... x -> -1

#include <iostream>
using namespace std;

const int maxn = 1e5 + 10;
int M, k, x;
char op;

int val[maxn], ne[maxn], head = -1, idx = 0;

void ins_h(int x) {
    idx++;
    val[idx] = x;
    ne[idx] = head;
    head = idx;
}

void ins(int k, int x) {
    idx++;
    val[idx] = x;
    ne[idx] = ne[k];
    ne[k] = idx;
}

void del(int k) { ne[k] = ne[ne[k]]; }

int main() {
    cin >> M;
    while (M--) {
        cin >> op;
        if (op == 'H') {
            cin >> x;
            ins_h(x);
        }
        if (op == 'D') {
            cin >> k;
            if (k == 0)
                head = ne[head];
            else
                del(k);
        }
        if (op == 'I') {
            cin >> k >> x;
            ins(k, x);
        }
    }

    for (int i = head; i != -1; i = ne[i]) printf("%d ", val[i]);

    return 0;
}

双向链表

 采用 1  idx idx idx ... idx 0 的写法,(毕竟不能访问l[-1]这种东西。。。）

从左往右获取即为

for (int i = r[0]; i != 1; i = r[i]) printf("%d ", val[i]);

#include <iostream>
using namespace std;
const int maxn = 4e5 + 10;

int val[maxn], l[maxn], r[maxn], M, x, k, idx;
string op;

void init() {
  r[0] = 1;
  l[1] = 0;
  idx = 1; // idx0 / idx1 function as endpoint of left / right
  // 0 idxl x idxlk k idlrk x x idxr 1
  //the direction of process depend on you insert in L or R
  // modify l -> modify r <-
}

void L(int x) {
  idx++;
  l[r[0]] = idx;
  l[idx] = 0;
  r[idx] = r[0];
  r[0] = idx;
  val[idx] = x;
}
void R(int x) {
  idx++;
  r[l[1]] = idx;
  r[idx] = 1;
  l[idx] = l[1];
  l[1] = idx;
  val[idx] = x;
}
void D(int k) {
  l[r[k]] = l[k];
  r[l[k]] = r[k];
}
void IL(int k, int x) {
  idx++;
  l[idx] = l[k];
  l[k] = idx;
  r[idx] = k;
  r[l[idx]] = idx;
  val[idx] = x;
}
void IR(int k, int x) {
  idx++;
  r[idx] = r[k];
  r[k] = idx;
  l[idx] = k;
  l[r[idx]] = idx;
  val[idx] = x;
}

int main() {
  init();
  cin >> M;
  while (M--) {
    cin >> op;
    if (op == "L") {
      cin >> x;
      L(x);
    }
    if (op == "R") {
      cin >> x;
      R(x);
    }
    if (op == "D") {
      cin >> k;
      D(k + 1);
    }
    if (op == "IL") {
      cin >> k >> x;
      IL(k + 1, x);
    }
    if (op == "IR") {
      cin >> k >> x;
      IR(k + 1, x);
    }
  }

  for (int i = r[0]; i != 1; i = r[i]) printf("%d ", val[i]);
  return 0;
}