Unique

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题目

实现类型版本的 Lodash.uniq 方法, Unique<T> 接收数组类型 T, 返回去重后的数组类型.

type Res = Unique<[1, 1, 2, 2, 3, 3]>; // expected to be [1, 2, 3]
type Res1 = Unique<[1, 2, 3, 4, 4, 5, 6, 7]>; // expected to be [1, 2, 3, 4, 5, 6, 7]
type Res2 = Unique<[1, "a", 2, "b", 2, "a"]>; // expected to be [1, "a", 2, "b"]
type Res3 = Unique<[string, number, 1, "a", 1, string, 2, "b", 2, number]>; // expected to be [string, number, 1, "a", 2, "b"]
type Res4 = Unique<[unknown, unknown, any, any, never, never]>; // expected to be [unknown, any, never]

解答

type Includes<T, U> = U extends [infer F, ...infer Rest]
  ? Equal<F, T> extends true
    ? true
    : Includes<T, Rest>
  : false;

type Unique<T, U extends any[] = []> = T extends [infer R, ...infer F]
  ? Includes<R, U> extends true
    ? Unique<F, [...U]>
    : Unique<F, [...U, R]>
  : U;

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这题需要借助辅助数组，通过递归的方式，依次把数组中没有的内容塞进去

Includes<R, U> extends true
      ? Unique<F, [...U]>
      : Unique<F, [...U, R]>

这里就是递归的核心逻辑，如果数组 U 中不包含该元素，就塞进去，有就不塞

一开始判断数组中是否存在该元素，采用的是 R extends U[number] 但是发现有很多 case 没有考虑到

因此需要实现一个 Includes 方法来判断是否有该值，这个方法也是常规的递归实现，不多赘述

// 答案
type Includes<T, U> = U extends [infer F, ...infer Rest]
  ? Equal<F, T> extends true
    ? true
    : Includes<T, Rest>
  : false;

type Unique<T, U extends any[] = []> = T extends [infer R, ...infer F]
  ? Includes<R, U> extends true
    ? Unique<F, [...U]>
    : Unique<F, [...U, R]>
  : U;