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Easy
JavaScript

中文文档

Description

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:

  • calls is a valid JSON array
  • 1 <= calls.length <= 10
  • 1 <= calls[i].length <= 100
  • 2 <= JSON.stringify(calls).length <= 1000

Solutions

Solution 1

TypeScript

type JSONValue = null | boolean | number | string | JSONValue[] | { [key: string]: JSONValue };
type OnceFn = (...args: JSONValue[]) => JSONValue | undefined;

function once(fn: Function): OnceFn {
    let called = false;
    return function (...args) {
        if (!called) {
            called = 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
 */

JavaScript

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