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(null,2,3); // 6
一次fn(null,3,6); // 未定义,fn 未被调用
**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:**
const oncefn = once(fn);
一次fn(null,7,4); // 140
一次fn(null,3,6); // 未定义,fn 未被调用
一次fn(null,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) => (a + b + c) * let onceFn = once(fn) * * onceFn(null,2,3); // 6 * onceFn(null,3,6); // returns undefined without calling fn */