【Q655】实现 intersection，取数组交集

[shfshanyue]

类似 lodash.intersection，有以下测试用例

//=> [2]
intersection([2, 1], [2, 3]);

//=> [1, 2]
intersection([1, 2, 2], [1, 2, 2])

//=> [1, 2]
intersection([1, 2, 2], [1, 2, 2], [1, 2])

[shfshanyue]

const intersection = (...list) => {
  const result = list.reduce((x, y) => x.filter(i => y.includes(i)))
  return [...new Set(result)]
}

[Kiera569]

const intersection = (...list) => list.reduce((a, b) => [...new Set(a.filter(item => b.includes(item)))])

[heretic-G]

function intersection (...args) {
    if (args.length === 0) return []
    for (let i = 0; i < args.length; i++) {
        if (!Array.isArray(args[i])) {
            args[i] = [args[i]]
        }
    }
    if (args.length === 1) return [...new Set(args[0])]
    let index = 1
    let sameArr = args[0]
    while (index < args.length) {
        let tempArr = []
        for (let i = 0;i < args[index].length; i++) {
            if (sameArr.includes(args[index][i])) {
                tempArr.push(args[index][i])
            }
        }
        sameArr = tempArr
        if (sameArr.length === 0) return []
        index += 1
    }
    return [...new Set(sameArr)]
}

[Asarua]

const intersection = (...args) => [...new Set(args.reduce((prev, next) => prev.filter(v => next.includes(v))))]

emm，绕了一圈绕回来了

[shfshanyue]

@Asarua 去重一下，看第二个示例，而且还有可能是多个数组呀，我去补一下测试用例

[Asarua]

@Asarua 去重一下，看第二个示例

已改

[haotie1990]

function intersection() {
  const arrays = [].slice.call(arguments, 0);
  const result = arrays.reduce(function(acc, arr){
    return acc.filter((v) => arr.indexOf(v) !== -1);
  });
  return result.reduce((acc, c) => {
    if (acc.indexOf(c) === -1) {
      acc.push(c);
    }
    return acc;
  }, []);
}

[rujptw]

// 取数组交集

/**
 * 
 * 
 * @param  {...any} arr 剩余数组
 * @returns 
 * 思路:
 * 1.设置两个数组，一个存放未重复的数组元素，一个存放已重复的数组元素
 * 2.在总的数组的循环中，如果数组元素在未重复数组中，又出现一次，则将其推入已重复数组
 * 3.使用Set,去除多余的元素，使用Array.from将Set实例转化为数组，并返回
 */
function intersection(...arr){
  let filterArr = [];
  let duplicateArr = []
  let sumArr = [].concat(...arr)
  sumArr.forEach(element => {
    if(!filterArr.includes(element)){
      filterArr.push(element)
    }else{
      duplicateArr.push(element)
    }
  });
  return Array.from(new Set(duplicateArr));
}

//=> [2]
console.log('intersection([2, 1], [2, 3]);: ', intersection([2, 1], [2, 3]));

//=> [1, 2]
console.log('intersection([1, 2, 2], [1, 2, 2]): ', intersection([1, 2, 2], [1, 2, 2]));

//=> [1, 2]
console.log('intersection([1, 2, 2], [1, 2, 2], [1, 2]): ', intersection([1, 2, 2], [1, 2, 2], [1, 2]));

[Hazel-Lin]

const intersection = (...arr) => [...new Set(arr.reduce((pre, cur) => pre.filter(item => cur.includes(item))))]
// [ 2, 3 ]
console.log(intersection([2, 3, 3], [2, 2, 3, 1], [2, 3, 1, 5]))