# 实现 intersection，取数组交集
更多描述
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]);
Issue
欢迎在 Gtihub Issue 中回答此问题: Issue 673 (opens new window)
Author
const intersection = (...list) => {
const result = list.reduce((x, y) => x.filter((i) => y.includes(i)));
return [...new Set(result)];
};
Author
const intersection = (...list) =>
list.reduce((a, b) => [...new Set(a.filter((item) => b.includes(item)))]);
Author
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)];
}
Author
const intersection = (...args) => [
...new Set(args.reduce((prev, next) => prev.filter((v) => next.includes(v)))),
];
emm，绕了一圈绕回来了
Author
@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
Author
@Asarua 去重一下，看第二个示例
已改
Author
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;
}, []);
}