数组去重，并将最大值放在最前面，最小值放在最后面，要求时间复杂度为 O(n)

1.js

const stat = (data) => {
  if (!Array.isArray(data) || data.length < 2) return data;
  let r = [], max = 0, min = 0, m = {}, vmax = data[0], vmin = data[0];
  data.forEach(v => {
    if (m[v]) return;
    m[v] = true;
    r.push(v);
    if (v > vmax) {
      vmax = v;
      max = r.length - 1;
    }
    if (v < vmin) {
      vmin = v;
      min = r.length - 1;
    }
  });
  const swap = (m, n) => {
    let temp = r[m];
    r[m] = r[n];
    r[n] = temp;
  };
  swap(max, 0);
  swap(min || max, r.length - 1);
  return r;
};

2.js

const stat = (data) => {
  if (!Array.isArray(data) || data.length < 2) return data;
  let r = [], max = 0, min = 0, m = {}, vmax = data[0], vmin = data[0];
  data.forEach(v => {
    if (m[v]) return;
    m[v] = true;
    r.push(v);
    if (v > vmax) {
      vmax = v;
      max = r.length - 1;
    }
    if (v < vmin) {
      vmin = v;
      min = r.length - 1;
    }
  });
  r.splice(max, 1);
  r.splice(min - (min > max), 1);
  r.unshift(vmax);
  r.push(vmin);
  return r;
};

3.js

const stat = (data) => {
  if (!Array.isArray(data) || data.length < 2) return data;
  let r = [], m = {}, vmax = data[0], vmin = data[0];
  data.forEach(v => {
    if (m[v]) return;
    m[v] = true;
    if (v > vmax) {
      r.unshift(vmax = v);
    } else if (v < vmin) {
      r.push(vmin = v);
    } else {
      if (r.length) {
        let t = r.pop();
        t < v ? r.push(v, t) : r.push(t, v);
      } else {
        r.push(v);
      }
    }
  });
  return r;
};