給定一個函數(shù) fn，返回一個與原始函數(shù)相同的新函數(shù)，除了它確保 fn 最多被調(diào)用一次。
第一次調(diào)用返回的函數(shù)時，它應(yīng)該返回與 fn 相同的結(jié)果。
隨后每次調(diào)用它時，它都應(yīng)該返回未定義。
示例1：
輸入：
fn = (a,b,c) => (a + b + c), 調(diào)用 = [[1,2,3],[2,3,6]]
輸出：
**explanation:**
登錄后復(fù)制
const oncefn = once(fn);
一次fn(1,2,3); // 6
一次fn(2,3,6); // 未定義，fn 未被調(diào)用
**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:**
登錄后復(fù)制
const oncefn = once(fn);
一次fn(5,7,4); // 140
一次fn(2,3,6); // 未定義，fn 未被調(diào)用
一次fn(4,6,8); // 未定義，fn 未被調(diào)用
**constraints:** `calls` is a valid json array
登錄后復(fù)制
1
1
2
*Solution*
In this case, we are required to create a higher-order function(a function that returns another function) [Read more about high-order functions here](https://www.freecodecamp.org/news/higher-order-functions-explained/#:~:text=JavaScript%20offers%20a%20powerful%20feature,even%20return%20functions%20as%20results.)
We should make sure that the original function `fn` is only called once regardless of how many times the second function is called.
If the function fn has been not called, we should call the function `fn` with the provided arguments `args`. Else, we should return `undefined`
_Code solution_
``` sh
/**
* @param {Function} fn
* @return {Function}
*/
var once = function (fn) {
// if function === called return undefined,
// else call fn with provide arguments
let executed = false;
let result;
return function (...args) {
if (!executed) {
executed = true;
result = fn(...args);
return result;
} else {
return undefined;
}
}
};
/**
* let fn = (a,b,c) => (a + b + c)
* let onceFn = once(fn)
*
* onceFn(1,2,3); // 6
* onceFn(2,3,6); // returns undefined without calling fn
*/
登錄后復(fù)制
