Question

2632. Curry

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Description

Given a function fn, return a curried version of that function.

A curried function is a function that accepts fewer or an equal number of parameters as the original function and returns either another curried function or the same value the original function would have returned.

In practical terms, if you called the original function like sum(1,2,3), you would call the curried version like csum(1)(2)(3), csum(1)(2,3), csum(1,2)(3), or csum(1,2,3). All these methods of calling the curried function should return the same value as the original.

 

Example 1:

Input: 
fn = function sum(a, b, c) { return a + b + c; }
inputs = [[1],[2],[3]]
Output: 6
Explanation:
The code being executed is:
const curriedSum = curry(fn);
curriedSum(1)(2)(3) === 6;
curriedSum(1)(2)(3) should return the same value as sum(1, 2, 3).

Example 2:

Input:
fn = function sum(a, b, c) { return a + b + c; }
inputs = [[1,2],[3]]
Output: 6
Explanation:
curriedSum(1, 2)(3) should return the same value as sum(1, 2, 3).

Example 3:

Input:
fn = function sum(a, b, c) { return a + b + c; }
inputs = [[],[],[1,2,3]]
Output: 6
Explanation:
You should be able to pass the parameters in any way, including all at once or none at all.
curriedSum()()(1, 2, 3) should return the same value as sum(1, 2, 3).

Example 4:

Input:
fn = function life() { return 42; }
inputs = [[]]
Output: 42
Explanation:
currying a function that accepts zero parameters should effectively do nothing.
curriedLife() === 42

 

Constraints:

• 1 <= inputs.length <= 1000
• 0 <= inputs[i][j] <= 105
• 0 <= fn.length <= 1000
• inputs.flat().length == fn.length
• function parameters explicitly defined
• If fn.length > 0 then the last array in inputs is not empty
• If fn.length === 0 then inputs.length === 1 

Solutions

Solution 1

function curry(fn: Function): Function {
  return function curried(...args) {
    if (args.length >= fn.length) {
      return fn(...args);
    }
    return (...nextArgs) => curried(...args, ...nextArgs);
  };
}

/**
 * function sum(a, b) { return a + b; }
 * const csum = curry(sum);
 * csum(1)(2) // 3
 */