Allow One Function Call 只呼叫一次有回傳的fn

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.

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

解法:

這是講一個函式傳入另一個函式，這個函式要控管他呼叫只能在第一次有回傳結果後續都是undefined

type Fn = (...args: any[]) => any;

const once = <T extends Fn>(fn: T): ((...args: Parameters<T>) => ReturnType<T> | undefined) => {
  let isCall = false;
  return (...args) => {
    if(isCall){
      return undefined;
    }else {
      isCall = true;
      return fn(...args);
    }
  }
}