1. 数组内元素为一般数据类型

const arr1 = [1, 2, 3, 4, 5, 6, 7];
const arr2 = [11, 2, 33, 4, 55, 6, 77, 6, 6];

// arr1和arr2交集预计输出 [2, 4, 6]

交集代表两个数组中共同具有且不重复的值，因此我们可以借助 ES6 中 Set 值不重复的特性来操作：

/**
* @param { Array } a
* @param { Array } b
* @return { Array }
*/
function intersect(a, b) {
  const setA = new Set(a);
  const setB = new Set(b);

  const intersection = [...setA].filter((item) => setB.has(item));

  return a.filter((item) => intersection.includes(item));
}

intersect(arr1, arr2); // [2, 4, 6]

2. 数组内元素为复杂数据类型

const arr1 = [
  {
    name: "张三",
    age: 18,
    job: "门卫",
  },
  {
    name: "李四",
    age: 28,
    job: "外卖小哥",
  },
  {
    name: "老哥",
    age: 38,
    job: "快递员",
  },
  {
    name: "大炮",
    age: 35,
    job: "电工",
  },
];
const arr2 = [
  {
    name: "铁柱",
    age: 35,
    job: "司机",
  },
  {
    name: "狗蛋",
    age: 22,
    job: "网红",
  },
  {
    name: "老哥",
    age: 38,
    job: "快递员",
  },
  {
    name: "阿牛",
    age: 35,
    job: "厨师",
  },
];
// 已知name为唯一标识，按name取交集，预计输出结果为[{name: "老哥", age: 38, job: "快递员"}]

实现也只需要在原本的函数上稍加修改即可

/**
* @param { Array } a
* @param { Array } b
* @param { String } prop
* @return { Array }
*/
function intersectByProp(a, b, prop) {
  const setA = new Set(a.map((item) => item[prop]));
  const setB = new Set(b.map((item) => item[prop]));

  const intersection = [...setA].filter((item) => setB.has(item));

  return a.filter((item) => intersection.includes(item[prop]));
}

intersectByProp(arr1, arr2); // [{name: "老哥", age: 38, job: "快递员"}]