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:**登录后复制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) => (a * b * c), calls = [[5,7,4],[2,3,6],[4,6,8]]```**output:** ```[{"calls":1,"value":140}]```**explanation:**登录后复制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登录后复制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 */登录后复制 以上就是Leetcode #允许一个函数调用的详细内容，更多请关注我的其它相关文章！