2666. Allow One Function Call

⭐

題目敘述

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

解題思路

Solution

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

/**
 * 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
 */

function once<T extends (...args: any[]) => any>(fn: T):
    ((...args: Parameters<T>) => ReturnType<T> | undefined) {
    let isCall: Boolean = false;
    return function (...args) {
        if(!isCall){
            isCall = true;
            return fn(...args);
        }
    };
}

/**
 * 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
 */