如何返回此数组中不重复的唯一值? [英] How to return the only value that is not repeated in this Array?

问题描述

我有一个关于Javascript的问题.

I have a question in Javascript.

这是以下数组:[1,0,0,0,0,0,0]

我想返回不重复的唯一值，即1.

I would like to return the only value that does not repeat, that is, 1.

有什么建议吗?

我有这个:

var result = arr.filter(x => arr.indexOf(x) !== 0);

推荐答案

使用indexOf和lastIndexOf

您可以比较indexOf和lastIndexOf和filter()

Using indexOf and lastIndexOf

You can compare indexOf and lastIndexOf and filter()

let arr = [1,0,0,0,0,0,0];
let res = arr.filter(x => arr.indexOf(x) === arr.lastIndexOf(x));

console.log(res)

如果只需要第一个元素，则可以使用find()

If you only want the first element you can use find()

let arr = [1,0,0,0,0,0,0];
let res = arr.find(x => arr.indexOf(x) === arr.lastIndexOf(x));

console.log(res)

您可以使用Set()删除重复项，然后在其上使用filter().

You can remove duplicates using Set() and then use filter() on it.

let arr = [1,0,0,0,0,0,0];
let res = [...new Set(arr)].filter(x => arr.indexOf(x) === arr.lastIndexOf(x));

console.log(res)

let arr = [1,0,0,0,0,0,0];
let res = arr.filter(x => arr.filter(a => a === x).length === 1);
console.log(res)

关于时间复杂度，这是最好的.

This one is best regarding time complexity.

let arr = [1,0,0,0,0,0,0];
let obj = arr.reduce((ac,a) => (ac[a] = ac[a] + 1 || 1,ac),{});
let res = Object.keys(obj).filter(x => obj[x] === 1).map(x => +x || x);

console.log(res)

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