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README_EN.md
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| true | Easy | https://github.com/doocs/leetcode/edit/main/solution/2600-2699/2666.Allow%20One%20Function%20Call/README_EN.md |
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2666. Allow One Function Call
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:
callsis a valid JSON array1 <= calls.length <= 101 <= calls[i].length <= 1002 <= 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
*/