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Chunk

2022-04-22 08

题目来源(type-challenges)

问题 

type case1 = Chunk<[], 1> //  []
type case2 = Chunk<[1, 2, 3], 1> //  [[1], [2], [3]]
type case3 = Chunk<[1, 2, 3], 2> //  [[1, 2], [3]]
type case4 = Chunk<[1, 2, 3, 4], 2> //  [[1, 2], [3, 4]]
type case5 = Chunk<[1, 2, 3, 4], 5> //  [[1, 2, 3, 4]]
type case6 = Chunk<[1, true, 2, false], 2> //  [[1, true], [2, false]]

解答 

type Chunk<T extends any[], N extends number, S extends any[] = []> = 
  S['length'] extends N
    ? [S, ...Chunk<T, N>]
    : T extends [infer F, ...infer R]
      ? Chunk<R, N, [...S, F]>
      : S extends [] ? S : [S]

拆分

  1. 首先判断分割数组的长度是否和目标长度相等
  2. 如果相等则返回当前分割好的数组加上后续需要递归处理的数组
  3. 然后拆分数组，如果可以拆分则递归处理
  4. 如果不能拆分，看数组是否为空，如果为空则返回，如果不为空则标志数组就剩下一个，直接进行包裹处理

执行过程 

type case5 = Chunk<[1, 2, 3, 4], 5> //  [[1, 2, 3, 4]]
TNSS['length']infer F...infer Rresult
[1, 2, 3, 4]5[]01[2, 3, 4]Chunk<[2, 3, 4], 5, [1]>
[2, 3, 4]5[1]12[3, 4]Chunk<[3, 4], 5, [1, 2]>
[3, 4]5[1, 2]23[4]Chunk<[4], 5, [1, 2, 3]>
[4]5[1, 2, 3]34[]Chunk<[], 5, [1, 2, 3, 4]>
[]5[1, 2, 3, 4]4nevernever[S]
[[1, 2, 3, 4]]
type case6 = Chunk<[1, true, 2, false], 2> //  [[1, true], [2, false]]
TNSS['length']infer F...infer Rresult
[1, true, 2, false]2[]01[true, 2, false]Chunk<[true, 2, false], 2, [1]>
[true, 2, false]2[1]1true[ 2, false]Chunk<[2, false], 2, [1, true]>
[2, false]2[1, true]2nevernever[S, ...Chunk<[2, false], 2>]
[2, false]2[]02[false]Chunk<[false], 2, [2]>
[false]2[2]1false[]Chunk<[], 2, [2, false]>
[]2[2, false]2nevernever[S, ...Chunk<[], 2>]
[]2[]0neverneverS
cd ..