[Typescript] 32. Medium - Permutation
作者:互联网
Implement permutation type that transforms union types into the array that includes permutations of unions.
type perm = Permutation<'A' | 'B' | 'C'>; // ['A', 'B', 'C'] | ['A', 'C', 'B'] | ['B', 'A', 'C'] | ['B', 'C', 'A'] | ['C', 'A', 'B'] | ['C', 'B', 'A']
Test cases:
/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'
type cases = [
Expect<Equal<Permutation<'A'>, ['A']>>,
Expect<Equal<Permutation<'A' | 'B' | 'C'>, ['A', 'B', 'C'] | ['A', 'C', 'B'] | ['B', 'A', 'C'] | ['B', 'C', 'A'] | ['C', 'A', 'B'] | ['C', 'B', 'A']>>,
Expect<Equal<Permutation<'B' | 'A' | 'C'>, ['A', 'B', 'C'] | ['A', 'C', 'B'] | ['B', 'A', 'C'] | ['B', 'C', 'A'] | ['C', 'A', 'B'] | ['C', 'B', 'A']>>,
Expect<Equal<Permutation<boolean>, [false, true] | [true, false]>>,
Expect<Equal<Permutation<never>, []>>,
]
Let's first solve the case:
Expect<Equal<Permutation<never>, []>>
, first idea would be:
type Permutation<T> = T extends never ? [] : [T];
type x = Permutation<never> // x is never
It doesn't work out, something nothing extends never, but if we put never into [never]
:
type Permutation<T> = [T] extends [never] ? [] : [T];
type x = Permutation<never> // x is []
Then let's see:
Expect<Equal<Permutation<boolean>, [false, true] | [true, false]>>,
, first try
type Permutation<T> = T extends boolean ? T : [];
type x = Permutation<boolean> // x is boolean
But if we put [T]
type Permutation<T> = T extends boolean ? [T] : [];
type x = Permutation<boolean> // x= [false] | [true]
So, boolean can be infer into false or true, that's great
Next step:
type Permutation<T, U = T> = [T] extends [never]
? []
: U extends infer X
? [X]
: []
type x =Permutation<'A' | 'B' | 'C'> // x = [A] | [B] | [C]
Final:
type Permutation<T, U = T> = [T] extends [never]
? []
: U extends infer X
? [X, ...Permutation<Exclude<T, X>>]
: []
标签:Typescript,false,32,never,Medium,extends,Expect,Permutation,type 来源: https://www.cnblogs.com/Answer1215/p/16695171.html