[Function, Closure] Allow One Function Call (함수 한 번만 호출하기)

https://leetcode.com/problems/allow-one-function-call/description/

Allow One Function Call - LeetCode

 

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

 

solution

/**
 * @param {Function} fn
 * @return {Function}
 */
var once = function(fn) {
    let isCall = true;
    return function(...args){
        if(isCall) {
            isCall = false;
            return fn(...args);
        } 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
 */