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Alex March
Asked:2023-01-08 05:02:33 +0000 UTC2023-01-08 05:02:33 +0000 UTC

js找到多个对象的交集（大于2）

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帮JS找几个对象的交集（2个以上） 比如找2个对象的交集有这么一段代码（下） 但是告诉我是不是对象很多？可能通过递归，但我找不到解决方案。

const intersection = (firstObject, secondObject) => {
    const newObj = {};
    for (let key in firstObject) { 
        if (key in secondObject && firstObject[key] === secondObject[key]) {
            newObj[key] = firstObject[key];
        }
    }           
    return newObj;
};
const obj1 = { a: 1, b: 2, c: 3 };
const obj2 = { b: 2, a: 2, c: 3 };
console.log(intersection(obj1, obj2));

javascript

1 个回答

  1. Best Answer

    不需要递归，依次寻找对象的交集即可：

    const objects1 = [
  { a: 1, b: 2, c: 3 },
  { a: 1, b: 2, c: 3 },
  { a: 1, b: 2, c: 3 },
  { a: 1, b: 2, c: 3 },
  { a: 1, b: 2, c: 3 }
];

const objects2 = [
  { a: 1, b: 6, c: 11 },
  { a: 2, b: 7, c: 12 },
  { a: 3, b: 8, c: 13 },
  { a: 4, b: 9, c: 14 },
  { a: 5, b: 10, c: 15 }
];


const intersectionOfTwoObjects = (firstObject, secondObject) => {
    const newObj = {};
    
    for (const key in firstObject) { 
        if (key in secondObject && firstObject[key] === secondObject[key]) {
            newObj[key] = firstObject[key];
        }
    }          
    
    return newObj;
};


const intersection = (objects) => {
  let intersectionObj = objects[0] ?? {};
  
  for (let i = 1; i < objects.length; ++i) {
    // check if intersectionObj is empty object
    if (Object.keys(intersectionObj).length === 0) return intersectionObj;
  
    intersectionObj = intersectionOfTwoObjects(intersectionObj, objects[i]);
  }
  
  return intersectionObj;
}

console.log(intersection(objects1));
console.log(intersection(objects2));

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