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「重学TS 2.0 」TS 练习题第二十七题
实现一个 Reverse 工具类型，用于对元组类型中元素的位置颠倒，并返回该数组。元组的第一个元素会变成最后一个，最后一个元素变成第一个。
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = // 你的实现代码
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
请在下面评论你的答案。
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = T extends [infer A,... infer B]? Reverse<B,[A,...R]> : R
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
type Reverse<
T extends Array<any>
> = T extends [infer First, ...infer Rest] ? [...Reverse<Rest>, First] : []
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
type Reverse<T extends Array<any>> = T extends [infer First, ...infer Rest]
? [...Reverse<Rest>, First]
: [];
type R0 = Reverse<[]>; // []
type R1 = Reverse<[1, 2, 3]>; // [3, 2, 1]
type Push<T extends any[], V> = [...T, V];
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = T extends [infer A, ...infer K] ? K extends [] ? Push<R, A> : Push<Reverse<K, R>, A> : []
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
向下递归时，变换一下两个 infer 变量的位置即可
type Reverse<T extends any[]> = T extends [infer R1, ...infer R2] ? [...Reverse<R2>, R1] : []
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = T extends [infer F, ...infer B] ? Reverse<B, [F, ...R]> : R// Reverse<B, [F, ...R]>就将前面的类型放置于后面
type R0 = Reverse<[]> // []
type R1 = Reverse<[1,2,3]> // [3,2,1]
题解
这里采用递归方式，每次递归都把第一项 First 放在最后，并把递归结果展开：
type Reverse<T extends Array<any>> =
T extends [infer First, ...infer Rest]
? [...Reverse<Rest>, First]
: [];
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
type R2 = Reverse<[1, 2, 3, 4, 5]> // [5, 4, 3, 2, 1]
拓展
本题还可以引入一个空数组暂存结果：
type Reverse<
T extends Array<any>,
A extends Array<any> = []
> = T extends [infer First, ...infer Rest] ? Reverse<Rest, [First, ...A]> : A;
第一次执行的时候，把 T 分成第一项 First 和剩余的 Rest，每次递归的时候，把 First 和暂存数组 A 合并，作为递归的第二个参数，即上一步翻转的结果，到最后返回暂存数组 A 即可。
类似的还可以这么处理：
type Reverse<
T extends Array<any>,
A extends Array<any> = []
> = T extends [...infer Rest, infer Last] ?
A['length'] extends T['length'] ? A :
Reverse<Rest, [...A, Last]> : A;
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
type R2 = Reverse<[1, 2, 3, 4, 5]> // [5, 4, 3, 2, 1]
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = T extends [infer A, ...infer B]
? [...Reverse<B>, A]
: [];
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3]> // [3, 2, 1]
这类型的问题，着眼点都在数组的形状上。
type Reverse<T extends Array<any>, R extends Array<any> = []> = T extends [...infer A, infer B] ? [B, ...Reverse<A>] : [];
type R0 = Reverse<[]>; // []
type R1 = Reverse<[1, 2, 3]>; // [3, 2, 1]
type Reverser<
T extends Array<any>,
R extends Array<any> = []
> = T extends [infer F, ...infer Rest]
? Reverser<Rest, [F,...R]>
: R
type R0 = Reverser<[]>
type R1 = Reverser<[1,2,3,4,5]>
type R0 = Reverse<[]> // []
type R1 = Reverse<[1, 2, 3, 4, 5]> // [3, 2, 1]
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> =
T extends [infer First, ...infer Rest]
? Reverse<Rest, [First, ...R]>
: R
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = T extends [infer First, ...infer Rest] ? Reverse<Rest, [First, ...R]> : R
type Reverse<
T extends Array<any>,
R extends Array<any> = []
> = T extends [] ? R : (
T extends [infer First, ...infer Rest] ? Reverse<Rest, [First, ...R]> : R
)
type Reverse<
T extends Array<any>,
R extends any[] = [],
> = T extends [infer A, ...infer Rest]
? Reverse<[...Rest], [A, ...R]>
: R
