FlattenDepth

约 180 字小于 1 分钟

2026-02-11

题目

Recursively flatten array up to depth times.

For example:

type a = FlattenDepth<[1, 2, [3, 4], [[[5]]]], 2> // [1, 2, 3, 4, [5]]. flattern 2 times
type b = FlattenDepth<[1, 2, [3, 4], [[[5]]]]> // [1, 2, 3, 4, [[5]]]. Depth defaults to be 1

If the depth is provided, it's guaranteed to be positive integer.

解题思路

待补充

答案

type FlattenDepth = any

验证

import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<FlattenDepth<[]>, []>>,
  Expect<Equal<FlattenDepth<[1, 2, 3, 4]>, [1, 2, 3, 4]>>,
  Expect<Equal<FlattenDepth<[1, [2]]>, [1, 2]>>,
  Expect<Equal<FlattenDepth<[1, 2, [3, 4], [[[5]]]], 2>, [1, 2, 3, 4, [5]]>>,
  Expect<Equal<FlattenDepth<[1, 2, [3, 4], [[[5]]]]>, [1, 2, 3, 4, [[5]]]>>,
  Expect<Equal<FlattenDepth<[1, [2, [3, [4, [5]]]]], 3>, [1, 2, 3, 4, [5]]>>,
  Expect<Equal<FlattenDepth<[1, [2, [3, [4, [5]]]]], 19260817>, [1, 2, 3, 4, 5]>>,
]

参考

无