type Includes<T extends unknown[], Target extends unknown> = T extends [
infer Head,
...infer Tail,
]
? Equal<Head, Target> extends true
? true
: Includes<Tail, Target>
: false;
type FindEles<T extends any[], U extends any[] = [], O extends any[] = []> = T extends [infer F, ...infer Rest]
? Includes<[...Rest, ...O],F> extends true
? FindEles<Rest, U, [...O, F]>
: FindEles<Rest, [...U, F], [...O, F]>
: U
Solution by wendao-liu #35098
type FindEles<T extends any[], U extends any[] = []> = T extends [infer First, ...infer Rest]
? First extends (Rest[number] | U[number]) ? FindEles<Rest, [...U, First]> : [First, ...FindEles<Rest, U>]
: []
Solution by 2083335157 #34942
// 你的答案
type IncludesItem<T, U> = U extends U ? [U] extends [T] ? true : false : false
type FindEles<T extends any[], U extends any[] = []> = T extends [infer A, ...infer R] ?
IncludesItem<A, U[number] | R[number]> extends false ? [A, ...FindEles<R, U>] : FindEles<R, [...U, A]>
: T
Solution by heyuelan #34727
type FindEles<T extends unknown[], U = never> = T extends [infer F, ...infer R]
? F extends U | R[number]
? FindEles<R, U | F>
: [F, ...FindEles<R, U | F>]
: [];
Solution by yukicountry #34352
type FindEles<T extends any[], A extends any[] = []> = T extends [infer F, ...infer R]
? F extends A[number]
? FindEles<R, A>
: F extends R[number] // Expect<Equal<FindEles<[1, 2, number]>, [1, 2, number]>> not pass
? FindEles<R, [...A, F]>
: [F, ...FindEles<R, [...A, F]>]
: []
In order for Expect<Equal<FindEles<[1, 2, number]>, [1, 2, number]>> to pass, replace F extends R[number] with R extends [...infer _, F]/R extends [F, ...infer _].
type FindEles<T extends any[], A extends any[] = []> = T extends [infer F, ...infer R]
? F extends A[number]
? FindEles<R, A>
: R extends [...infer _, F]
? FindEles<R, [...A, F]>
: R extends [F, ...infer _]
? FindEles<R, [...A, F]>
: [F, ...FindEles<R, [...A, F]>]
: []
Solution by MAXLZ1 #34310
type FindEles<T extends any[], Finded extends any[] = [], Res extends any[] = []> = T extends [infer R, ...infer rest] ?
R extends [...Finded, ...rest][number] ? FindEles<rest, [...Finded, R], Res> : FindEles<rest, Finded, [...Res, R]> : Res
Solution by ouzexi #34114
type isEqual<F, S> = (<G>() => G extends F ? 1 : 2) extends (<G>() => G extends S ? 1 : 2) ? true : false
type Remove<T extends any[], S, Result extends any[] = []> = T extends [infer First, ...infer Rest]
? isEqual<First, S> extends true
? [...Result, ...Rest]
: Remove<Rest, S, [...Result, First]>
: Result
type FindEles<T extends any[], All extends any = never, Unique extends any[] = []> = T extends [infer First, ...infer Rest]
? First extends All
? FindEles<Rest, All, Remove<Unique, First>>
: FindEles<Rest, All | First, [...Unique, First]>
: Unique
Solution by PiligrimStas #33804
type FindEles<
T extends any[],
Arr extends any[] = [],
Ret extends any[] = []
> = T extends [infer Ele, ...infer Rest]
? Ele extends [...Rest, ...Arr][number]
? FindEles<Rest, [...Arr, Ele], Ret>
: FindEles<Rest, [...Arr], [...Ret, Ele]>
: Ret;
Solution by sunupupup #33447
type FindEles<T extends number[], ValKeys = {[K in keyof T & `${number}` as T[K]]: K}, R extends number[] = []>
= T extends [...infer Lead extends number[], infer Last extends keyof ValKeys & number]
? FindEles<Lead, ValKeys, ValKeys[Last] extends `${Lead['length']}` ? [Last, ...R] : R>
: R
Solution by teamchong #33352
// your answers
type FindEles<T extends any[]> = T extends [infer S, ...infer L]
? L extends S
? FindEles<L> : [S]
:[]
Solution by pea-sys #33351
type FindEles<T extends any[], VisitedItems extends unknown[] = [], UniqItems extends unknown[] = []> = T extends [infer First, ...infer Rest]
? First extends (VisitedItems[number] | Rest[number])
? FindEles<Rest, [...VisitedItems, First], UniqItems>
: FindEles<Rest, [...VisitedItems, First], [...UniqItems, First]>
: UniqItems
Solution by keyurparalkar #32726
type Filter<T extends any[], P, R extends any[] = []> = T["length"] extends 0
? R
: T[0] extends P
? Filter<T extends [any, ...infer U] ? U : never, P, [...R, T[0]]>
: Filter<T extends [any, ...infer U] ? U : never, P, R>;
type FindEles<T extends number[], R extends number[] = T, Once extends number = never, More extends number = never> =
R["length"] extends 0
? Filter<T, Once>
: R extends [infer F extends number, ...infer U extends number[]]
? F extends More
? FindEles<T, U, Once, More>
: F extends Once
? FindEles<T, U, Exclude<Once,F>, More | F>
: FindEles<T, U, Once | F, More>
: never;
Solution by vangie #32196
type FindEles<T extends readonly any[], Uniques extends readonly any[] = [], Seen extends readonly any[] = []> =
T extends [infer F, ...infer Tail] // does the list contains any element?
? F extends Seen[number] // the first element of the list is already seen?
? FindEles<Tail, Uniques, Seen>
: F extends Tail[number] // is the first element contained in the rest of the list?
? FindEles<Tail, Uniques, [...Seen, F]>
: FindEles<Tail, [...Uniques, F], [...Seen, F]>
: Uniques
I tried to make the generic types very self explicables
Solution by joyanedel #32157
Readable solution
type FindEles<T extends unknown[]> = FindElesInternal<T>;
type FindElesInternal<
Cur extends unknown[],
Prev extends unknown[] = [],
Acc extends unknown[] = []
> =
Cur extends [infer Head, ...infer Tail]
? Head extends [...Prev, ...Tail][number]
? FindElesInternal<Tail, [...Prev, Head], Acc>
: FindElesInternal<Tail, [...Prev, Head], [...Acc, Head]>
: Acc
;
Solution by sdrjs #31844
// 你的答案
type FindEles<T extends any[], P extends any[] = []> =
T extends [infer S, ...infer O] ? (S extends P[number] ? FindEles<O, [...P, S]> : (S extends O[number] ? FindEles<O, [...P, S]> : [S, ...FindEles<O, [...P, S]>])) : [];
Solution by tarotlwei #31746
type FindOnceElement<
T extends any[],
P extends any[] = [],
K extends any[] = []
> = T extends [infer F, ...infer R]
? FindOnceElement<
R,
[...P, F],
F extends [...P, ...R][number] ? K : [...K, F]
>
: K;
Solution by moonpoet #31027
type FindEles<T extends unknown[], R extends unknown[] = [], U extends unknown[] = []> = T extends [
infer F,
...infer Rest,
]
? F extends Rest[number] | U[number]
? FindEles<Rest, [...R], [...U, F]>
: FindEles<Rest, [...R, F], [...U]>
: R;
Solution by leejaehyup #30865
type FindEles<T extends any[], U extends any[] = [], O extends any[] = []> = T extends [infer A, ...infer B] ? A extends O[number] ? FindEles<B, U, [...O, A]> : A extends B[number] ? FindEles<B, U, [...O, A]> : FindEles<B, [...U, A], [...O, A]> : U
Solution by dreamluo-plus #30685
// 1.声明一个MyIncludes, 用来判断当前元素是否出现在其他元素中(MyIncludes可以根据场景来考虑是否要Equal)
// 2.Iterated用来收集遍历过的元素, List用来收集没出现过的原
// 3.解构第一个元素, 用他和其他的元素作对比, 其他元素就是Iterated + Other
// 4.如果没出现, 就放到List, 然后继续递归
type MyIncludes<T extends any[], K> = T extends [infer First, ...infer Other] ? First extends K ? true : MyIncludes<Other, K> : false
type FindEles<T extends any[], Iterated extends any[] = [], List extends any[] = []> = T extends [infer First, ...infer Other]
? MyIncludes<[...Iterated, ...Other], First> extends true
? FindEles<Other, [...Iterated, First], List>
: FindEles<Other, [...Iterated, First], [...List, First]>
: List
Solution by xiahouwei #30586
type ToString<T extends unknown[], Result extends string = ''> = T extends [
infer R extends number,
...infer Rest,
]
? ToString<Rest, `${Result}${R}`>
: Result
type FindEles<T extends unknown[], S = ToString<T>, Result extends unknown[] = []> = T extends [
infer R extends number,
...infer Rest,
]
? FindEles<Rest, S, S extends `${string}${R}${string}${R}${string}` ? Result : [...Result, R]>
: Result
先把数组转成字符串，再推断就简单了
Solution by zhangyu1818 #30274
Duplicates用于存储重复出现的元素，当F即不在T中也不再Duplicates中，则表示它只出现了一次
type FindEles<T extends any[], Duplicates extends any[] =[]> =
T extends [infer F, ...infer R]
? F extends R[number]
? FindEles<R, [...Duplicates, F]> : F extends Duplicates[number]
? FindEles<R, Duplicates> : [F, ...FindEles<R, Duplicates>]
:[]
Solution by sv-98-maxin #30127
type FindEles<
T extends any[],
P extends any[] = [],
R extends any[] = []
> = T extends [infer First, ...infer Rest]
? First extends [...P, ...Rest][number]
? FindEles<Rest, [...P, First], R>
: FindEles<Rest, P, [...R, First]>
: R;
Solution by idebbarh #29958
type FindEles<T extends unknown[], U = never> =
T extends [infer F, ...infer R]
? F extends U | R[number]
? FindEles<R, U | F>
: [F, ...FindEles<R, U>]
: [];
Solution by ttkien2125 #29795
type FindEles<T extends any[], Prevs extends any[] = []> = T extends [
infer Item,
...infer Rest
]
? Item extends Prevs[number] | Rest[number]
? FindEles<Rest, [...Prevs, Item]>
: [Item, ...FindEles<Rest, [...Prevs, Item]>]
: [];
Solution by DoubleWoodLin #28853
type In<T extends any[], U> = U extends T[number] ? true : false;
type Filter<T extends any[], U extends any, S extends any[] = []> = T extends [infer L, ...infer Rest]
? Filter<Rest, U, L extends U
? S
: [...S, L]>
: S
type FindEles<T extends any[], S extends any[] = []> = T extends [infer First, ...infer Rest]
? FindEles<Filter<T, First>, In<Rest, First> extends true
? S
: [...S, First]>
: S;
Solution by JohnhanLiu #28832
type ValueInArr<T extends any[], Val extends any> = T extends [infer First, ...infer Rest] ? First extends Val ? true : ValueInArr<Rest, Val> : false
type FindEles<T extends any[], Trash extends any[] = [], Acc extends any[] = []> = T extends [infer First, ...infer Rest] ? ValueInArr<[...Rest, ...Trash], First> extends true ? FindEles<Rest, [...Trash, First], Acc> : FindEles<Rest, Trash, [...Acc, First]> : Acc
Solution by julianmendonca #28813
type IsRepeat<T extends unknown[], I> = T extends [infer First, ...infer Rest]
? Equal<I, First> extends true
? true
: IsRepeat<Rest, I>
: false;
type FindEles<T extends unknown[], List extends unknown[] = [], Result extends unknown[] = []> = List['length'] extends T['length']
? Result
: T extends [infer First, ...infer Rest]
? IsRepeat<Rest, First> extends true
? FindEles<[...Rest, First], [...List, First], Result>
: FindEles<[...Rest, First], [...List, First], [...Result, First]>
: never;
Solution by NameWjp #28751
// 你的答案
type FindEles<
T extends any[],
K extends any[] = [],
U extends any[] = [],
O extends { [index: string]: any } = {},
> = T extends [infer A extends number | string, ...infer B]
? A extends B[number]
?
FindEles<
B,
[...K, 1],
U,
{ [P in keyof O | A]: P }
>
: unknown extends O[A] ?
FindEles<
B,
[...K, 1],
[...U, A],
O>
:
FindEles<
B,
[...K, 1],
U,
O
>
: U
Solution by xpbsm #28586
type FindEles<
Items extends unknown[],
ProcessedItems extends unknown[] = [],
UniqueItems extends unknown[] = []
> = Items extends [infer Head, ...infer Tail]
? FindEles<
Tail,
[...ProcessedItems, Head],
Head extends [...ProcessedItems, ...Tail][number]
? UniqueItems
: [...UniqueItems, Head]
>
: UniqueItems;
Solution by yevhenpavliuk #28416
type Include<T, U> = T extends [infer F, ...infer R]
? Equal<F, U> extends true
? true
: Include<R, U>
: false;
type FindEles<
T extends unknown[],
A extends unknown[] = [],
P extends unknown[] = []
> = T extends [infer F, ...infer R]
? Include<[...P, ...R], F> extends true
? FindEles<R, A, [...P, F]>
: FindEles<R, [...A, F], [...P, F]>
: A;
Solution by DoGukKim #28273