2666. 只允许一次函数调用
2666. 只允许一次函数调用
题目
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 array1 <= calls.length <= 10
1 <= calls[i].length <= 100
2 <= JSON.stringify(calls).length <= 1000
题目大意
给定一个函数 fn
,它返回一个新的函数,返回的函数与原始函数完全相同,只不过它确保 fn
最多被调用一次。
- 第一次调用返回的函数时,它应该返回与
fn
相同的结果。 - 第一次后的每次调用,它应该返回
undefined
。
提示:
calls
是一个有效的 JSON 数组1 <= calls.length <= 10
1 <= calls[i].length <= 100
2 <= JSON.stringify(calls).length <= 1000
解题思路
这道题的核心在于“函数只能被调用一次”的机制,这个设计模式非常适合一些需要控制函数调用次数的场景,比如防止多次执行某个动作。
可以利用闭包来实现,闭包使得我们能够在返回的函数中保持对原始函数调用状态的追踪,从而控制它只能被调用一次。
- 使用一个标志
called
来跟踪该函数是否已经被调用过,初始为false
。 - 返回的匿名函数,每次调用时会检查
called
是否为true
。如果为false
,则调用fn
,否则返回undefined
。
复杂度分析
- 时间复杂度:
O(1)
,每次调用时只检查标志变量并做出相应操作。 - 空间复杂度:
O(1)
,只存储一个布尔变量来跟踪调用状态。
代码
/**
* @param {Function} fn
* @return {Function}
*/
var once = function (fn) {
let called = false; // 用于跟踪函数是否已被调用
return function (...args) {
if (!called) {
called = true; // 第一次调用时,标记为已调用
return fn(...args); // 调用传入的函数,并传递参数
}
return undefined; // 如果已被调用,返回 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
*/