LeetCode 题解工作台
只允许一次函数调用
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 。 示例 1： 输入： fn = (a,b,c) => (a + b + c),…
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题型
2
代码语言
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简单 · Allow One Function Call core interview pattern
答案摘要
type JSONValue = null | boolean | number | string | JSONValue[] | { [key: string]: JSONValue }; type OnceFn = (...args: JSONValue[]) => JSONValue | undefined;
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题目描述
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。
示例 1：
输入：fn = (a,b,c) => (a + b + c), calls = [[1,2,3],[2,3,6]]
输出：[{"calls":1,"value":6}]
解释：
const onceFn = once(fn);
onceFn(1, 2, 3); // 6
onceFn(2, 3, 6); // undefined, fn 没有被调用
示例 2：
输入：fn = (a,b,c) => (a * b * c), calls = [[5,7,4],[2,3,6],[4,6,8]]
输出：[{"calls":1,"value":140}]
解释：
const onceFn = once(fn);
onceFn(5, 7, 4); // 140
onceFn(2, 3, 6); // undefined, fn 没有被调用
onceFn(4, 6, 8); // undefined, fn 没有被调用
提示：
calls是一个有效的 JSON 数组1 <= calls.length <= 101 <= calls[i].length <= 1002 <= JSON.stringify(calls).length <= 1000
解题思路
方法一
/**
* @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
*/
复杂度分析
| 指标 | 值 |
|---|---|
| 时间 | and space complexity depend on the function fn and the number of arguments passed. Overhead is minimal, with a single boolean flag and optional result storage, so operations are effectively O(1) per call. |
| 空间 | Depends on the final approach |
面试官常问的追问
外企场景- question_mark
Check if the candidate tracks function execution state correctly.
- question_mark
Observe if they handle extra calls without throwing errors.
- question_mark
Look for proper argument forwarding and result preservation in the wrapper.
常见陷阱
外企场景- error
Failing to store the result of the first call, causing loss of output for debugging.
- error
Not handling additional arguments correctly, breaking the function signature.
- error
Allowing multiple executions due to missing or misplaced execution flag.
进阶变体
外企场景- arrow_right_alt
Allow multiple calls but limit total executions to a specified count.
- arrow_right_alt
Reset the single-call state after a timeout, allowing one call per interval.
- arrow_right_alt
Track the last returned value and always return it instead of undefined for extra calls.