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# 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
# Solution
/** | |
* @param {Function} fn | |
* @return {Function} | |
*/ | |
var once = function(fn) { | |
let cnt = 0; | |
return function(...args){ | |
cnt++; | |
if (cnt == 1) 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 | |
*/ |
function once<T extends (...args: any[]) => any>(fn: T): | |
((...args: Parameters<T>) => ReturnType<T> | undefined) { | |
let cnt = 0; | |
return function (...args) { | |
cnt++; | |
if (cnt == 1) 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 | |
*/ |
