Leetcode #允许一个函数调用- 技术经验 -卓越飞翔博客

leetcode #允许一个函数调用

给定一个函数 fn，返回一个与原始函数相同的新函数，除了它确保 fn 最多被调用一次。

第一次调用返回的函数时，它应该返回与 fn 相同的结果。
随后每次调用它时，它都应该返回未定义。

示例1：

输入：

fn = (a,b,c) => (a + b + c), 调用 = [[1,2,3],[2,3,6]]

输出：



**explanation:**

const oncefn = once(fn);
一次fn(1,2,3); // 6
一次fn(2,3,6); // 未定义，fn 未被调用



**example 2:**

**input:**

 ```fn = (a,b,c) =&gt; (a * b * c), calls = [[5,7,4],[2,3,6],[4,6,8]]```


**output:**

 ```[{"calls":1,"value":140}]```


**explanation:**

const oncefn = once(fn);
一次fn(5,7,4); // 140
一次fn(2,3,6); // 未定义，fn 未被调用
一次fn(4,6,8); // 未定义，fn 未被调用




**constraints:**

`calls` is a valid json array

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) =&gt; (a + b + c)
 * let onceFn = once(fn)
 *
 * onceFn(1,2,3); // 6
 * onceFn(2,3,6); // returns undefined without calling fn
 */

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