2666. Allow One Function Call

tags: leetcode 30 days js challenge Easy

2666. Allow One Function Call

題目描述

Given a function fn, return a new function that is identical to the original function except that it ensures fn is called at most once.

• The first time the returned function is called, it should return the same result as fn.
• Every subsequent time it is called, it should return undefined.

範例

Example 1:

Input: fn = (a,b,c) => (a + b + c), calls = [[1,2,3],[2,3,6]]
Output: [{"calls":1,"value":6}]
Explanation:
const onceFn = once(fn);
onceFn(1, 2, 3); // 6
onceFn(2, 3, 6); // undefined, fn was not called

Example 2:

Input: fn = (a,b,c) => (a * b * c), calls = [[5,7,4],[2,3,6],[4,6,8]]
Output: [{"calls":1,"value":140}]
Explanation:
const onceFn = once(fn);
onceFn(5, 7, 4); // 140
onceFn(2, 3, 6); // undefined, fn was not called
onceFn(4, 6, 8); // undefined, fn was not called

Constraints:

• 1 <= calls.length <= 10
• 1 <= calls[i].length <= 100
• 2 <= JSON.stringify(calls).length <= 1000

解答

TypeScript

function once<T extends (...args: any[]) => any>(
  fn: T
): (...args: Parameters<T>) => ReturnType<T> | undefined {
  let hasBeenCalled = false;
  return (...args) => {
    if (hasBeenCalled) return;
    hasBeenCalled = true;
    return fn(...args);
  };
}

SheepFri, May 12, 2023

Reference

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