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 <= 101 <= calls[i].length <= 1002 <= JSON.stringify(calls).length <= 1000
Solution:
JavaScript
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// Author: Huahua /** * @param {Function} fn * @return {Function} */ var once = function(fn) { this.called = false; return function(...args){ if (this.called === false) { this.called = true; return fn(...args); } else { return undefined; } } }; |
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