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[Function, Closure] Allow One Function Call (함수 한 번만 호출하기) 본문
https://leetcode.com/problems/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:
- 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
*/
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