解决 TS 问题的最好办法就是多练,这次解读 type-challenges Medium 难度 63~68 题。
实现 Unique<T>
,对 T
去重:
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]
去重需要不断递归产生去重后结果,因此需要一个辅助变量 R
配合,并把 T
用 infer
逐一拆解,判断第一个字符是否在结果数组里,如果不在就塞进去:
type Unique<T, R extends any[] = []> = T extends [infer F, ...infer Rest] ? Includes<R, F> extends true ? Unique<Rest, R> : Unique<Rest, [...R, F]> : R
那么剩下的问题就是,如何判断一个对象是否出现在数组中,使用递归可以轻松完成:
type Includes<Arr, Value> = Arr extends [infer F, ...infer Rest] ? Equal<F, Value> extends true ? true : Includes<Rest, Value> : false
每次取首项,如果等于 Value
直接返回 true
,否则继续递归,如果数组递归结束(不构成 Arr extends [xxx]
的形式)说明递归完了还没有找到相等值,直接返回 false
。
把这两个函数组合一下就能轻松解决本题:
// 本题答案 type Unique<T, R extends any[] = []> = T extends [infer F, ...infer Rest] ? Includes<R, F> extends true ? Unique<Rest, R> : Unique<Rest, [...R, F]> : R type Includes<Arr, Value> = Arr extends [infer F, ...infer Rest] ? Equal<F, Value> extends true ? true : Includes<Rest, Value> : false
实现 MapTypes<T, R>
,根据对象 R
的描述来替换类型:
type StringToNumber = { mapFrom: string; // value of key which value is string mapTo: number; // will be transformed for number } MapTypes<{iWillBeANumberOneDay: string}, StringToNumber> // gives { iWillBeANumberOneDay: number; }
因为要返回一个新对象,所以我们使用 { [K in keyof T]: ... }
的形式描述结果对象。然后就要对 Value 类型进行判断了,为了防止 never
的作用,我们包一层数组进行判断:
type MapTypes<T, R extends { mapFrom: any; mapTo: any }> = { [K in keyof T]: [T[K]] extends [R['mapFrom']] ? R['mapTo'] : T[K] }
但这个解答还有一个 case 无法通过:
MapTypes<{iWillBeNumberOrDate: string}, StringToDate | StringToNumber> // gives { iWillBeNumberOrDate: number | Date; }
我们需要考虑到 Union 分发机制以及每次都要重新匹配一次是否命中 mapFrom
,因此需要抽一个函数:
type Transform<R extends { mapFrom: any; mapTo: any }, T> = R extends any ? T extends R['mapFrom'] ? R['mapTo'] : never : never
为什么要 R extends any
看似无意义的写法呢?原因是 R
是联合类型,这样可以触发分发机制,让每一个类型独立判断。所以最终答案就是:
// 本题答案 type MapTypes<T, R extends { mapFrom: any; mapTo: any }> = { [K in keyof T]: [T[K]] extends [R['mapFrom']] ? Transform<R, T[K]> : T[K] } type Transform<R extends { mapFrom: any; mapTo: any }, T> = R extends any ? T extends R['mapFrom'] ? R['mapTo'] : never : never
生成指定长度的 Tuple:
type result = ConstructTuple<2> // expect to be [unknown, unkonwn]
比较容易想到的办法是利用下标递归:
type ConstructTuple< L extends number, I extends number[] = [] > = I['length'] extends L ? [] : [unknown, ...ConstructTuple<L, [1, ...I]>]
但在如下测试用例会遇到递归长度过深的问题:
ConstructTuple<999> // Type instantiation is excessively deep and possibly infinite
一种解法是利用 minusOne 提到的 CountTo
方法快捷生成指定长度数组,把 1
替换为 unknown
即可:
// 本题答案 type ConstructTuple<L extends number> = CountTo<`${L}`> type CountTo< T extends string, Count extends unknown[] = [] > = T extends `${infer First}${infer Rest}` ? CountTo<Rest, N<Count>[keyof N & First]> : Count type N<T extends unknown[] = []> = { '0': [...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T] '1': [...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown] '2': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown ] '3': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown ] '4': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown, unknown ] '5': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown, unknown, unknown ] '6': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown, unknown, unknown, unknown ] '7': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown, unknown, unknown, unknown, unknown ] '8': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown ] '9': [ ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown ] }
实现 NumberRange<T, P>
,生成数字为从 T
到 P
的联合类型:
type result = NumberRange<2, 9> // | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
以 NumberRange<2, 9>
为例,我们需要实现 2
到 9
的递增递归,因此需要一个数组长度从 2
递增到 9
的辅助变量 U
,以及一个存储结果的辅助变量 R
:
type NumberRange<T, P, U extends any[] = 长度为 T 的数组, R>
所以我们先实现 LengthTo
函数,传入长度 N
,返回一个长度为 N
的数组:
type LengthTo<N extends number, R extends any[] = []> = R['length'] extends N ? R : LengthTo<N, [0, ...R]>
然后就是递归了:
// 本题答案 type NumberRange<T extends number, P extends number, U extends any[] = LengthTo<T>, R extends number = never> = U['length'] extends P ? ( R | U['length'] ) : ( NumberRange<T, P, [0, ...U], R | U['length']> )
R
的默认值为 never
非常重要,否则默认值为 any
,最终类型就会被放大为 any
。
实现 Combination<T>
:
// expected to be `"foo" | "bar" | "baz" | "foo bar" | "foo bar baz" | "foo baz" | "foo baz bar" | "bar foo" | "bar foo baz" | "bar baz" | "bar baz foo" | "baz foo" | "baz foo bar" | "baz bar" | "baz bar foo"` type Keys = Combination<['foo', 'bar', 'baz']>
本题和 AllCombination
类似:
type AllCombinations_ABC = AllCombinations<'ABC'> // should be '' | 'A' | 'B' | 'C' | 'AB' | 'AC' | 'BA' | 'BC' | 'CA' | 'CB' | 'ABC' | 'ACB' | 'BAC' | 'BCA' | 'CAB' | 'CBA'
还记得这题吗?我们要将字符串变成联合类型:
type StrToUnion<S> = S extends `${infer F}${infer R}` ? F | StrToUnion<R> : never
而本题 Combination
更简单,把数组转换为联合类型只需要 T[number]
。所以本题第一种组合解法是,将 AllCombinations
稍微改造下,再利用 Exclude
和 TrimRight
删除多余的空格:
// 本题答案 type AllCombinations<T extends string[], U extends string = T[number]> = [ U ] extends [never] ? '' : '' | { [K in U]: `${K} ${AllCombinations<never, Exclude<U, K>>}` }[U] type TrimRight<T extends string> = T extends `${infer R} ` ? TrimRight<R> : T type Combination<T extends string[]> = TrimRight<Exclude<AllCombinations<T>, ''>>
还有一种非常精彩的答案在此分析一下:
// 本题答案 type Combination<T extends string[], U = T[number], A = U> = U extends infer U extends string ? `${U} ${Combination<T, Exclude<A, U>>}` | U : never;
依然利用 T[number]
的特性将数组转成联合类型,再利用联合类型 extends
会分组的特性递归出结果。
之所以不会出现结尾出现多余的空格,是因为 U extends infer U extends string
这段判断已经杜绝了 U
消耗完的情况,如果消耗完会及时返回 never
,所以无需用 TrimRight
处理右侧多余的空格。
至于为什么要定义 A = U
,在前面章节已经介绍过了,因为联合类型 extends
过程中会进行分组,此时访问的 U
已经是具体类型了,但此时访问 A
还是原始的联合类型 U
。
实现 Subsequence<T>
输出所有可能的子序列:
type A = Subsequence<[1, 2]> // [] | [1] | [2] | [1, 2]
因为是返回数组的全排列,只要每次取第一项,与剩余项的递归构造出结果,|
上剩余项本身递归的结果就可以了:
// 本题答案 type Subsequence<T extends number[]> = T extends [infer F, ...infer R extends number[]] ? ( Subsequence<R> | [F, ...Subsequence<R>] ) : T
对全排列问题有两种经典解法:
本文作者:前端小毛
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