JavaScript 从浅入深 100 题训练计划--(22)数组交集 / 并集 / 差集

题目：实现数组的交集、并集、差集（对称差集）

intersection([1,2,3], [2,3,4]) // [2,3]
union([1,2,3], [2,3,4]) // [1,2,3,4]
difference([1,2,3], [2,3,4]) // [1]

 

const A = [1, 2, 3, 4, 5];
const B = [4, 5, 6, 7, 8];

function intersection(arr1, arr2) {
  const list = new Set(arr2);
  return arr1.filter(item => list.has(item))
}

function union(arr1, arr2) {
  return [...new Set([...arr1, ...arr2])]
}

// 不去重的版本
const result1 = A.filter(item => new Set(B).has(item));
// [4, 5]  ← 正好没重复，但如果 A = [1,1,2,4,5]，会返回 [1,1,4,5]

// 去重的版本（结果也是 Set）
function intersectionUnique(arr1, arr2) {
  const set2 = new Set(arr2);
  return [...new Set(arr1.filter(item => set2.has(item)))];
}


function difference(arr1, arr2) {
  const list = new Set(arr2);
  return arr1.filter(item => !list.has(item))
}

function symmetricDifference(arr1, arr2) {
  const list1 = new Set(arr1)
  const list2 = new Set(arr2)

  return [
    ...arr1.filter(item => !list2.has(item)),
    ...arr2.filter(item => !list1.has(item)),
  ]
}

class ArraySet {
  // 交集
  static intersection(...arrays) {
    if (arrays.length === 0) return [];
    return arrays.reduce((a, b) => a.filter(x => b.includes(x)));
  }

  // 并集
  static union(...arrays) {
    return [...new Set(arrays.flat())];
  }

  // 差集 (a - b)
  static difference(arr1, arr2) {
    const set2 = new Set(arr2);
    return arr1.filter(x => !set2.has(x));
  }

  // 对称差集
  static symmetricDifference(arr1, arr2) {
    return [
      ...arr1.filter(x => !arr2.includes(x)),
      ...arr2.filter(x => !arr1.includes(x))
    ];
  }

  // 判断是否子集
  static isSubset(arr1, arr2) {
    const set2 = new Set(arr2);
    return arr1.every(x => set2.has(x));
  }
}

console.log(intersection(A, B), 'key', union(A, B), result1, intersectionUnique(A, B), difference(A, B), difference(B, A), symmetricDifference(A, B))