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

 */