LeetCode 题解工作台
计数器 II
请你写一个函数 createCounter 。这个函数接收一个初始的整数值 init 。并返回一个包含三个函数的对象。 这三个函数是： increment() 将当前值加 1 并返回。 decrement() 将当前值减 1 并返回。 reset() 将当前值设置为 init 并返回。 示例 1： …
0
题型
1
代码语言
0
相关题
当前训练重点
简单 · Counter II core interview pattern
答案摘要
type ReturnObj = { increment: () => number;
Interview AiBoxInterview AiBox 实时 AI 助手，陪你讲清 Counter II core interview pattern 题型思路
题目描述
请你写一个函数 createCounter。这个函数接收一个初始的整数值 init。并返回一个包含三个函数的对象。
这三个函数是：
increment()将当前值加 1 并返回。decrement()将当前值减 1 并返回。reset()将当前值设置为init并返回。
示例 1：
输入：init = 5, calls = ["increment","reset","decrement"] 输出：[6,5,4] 解释： const counter = createCounter(5); counter.increment(); // 6 counter.reset(); // 5 counter.decrement(); // 4
示例 2：
输入：init = 0, calls = ["increment","increment","decrement","reset","reset"] 输出：[1,2,1,0,0] 解释： const counter = createCounter(0); counter.increment(); // 1 counter.increment(); // 2 counter.decrement(); // 1 counter.reset(); // 0 counter.reset(); // 0
提示：
-1000 <= init <= 10000 <= calls.length <= 1000calls[i]是 “increment”、“decrement” 和 “reset” 中的一个
解题思路
方法一
type ReturnObj = {
increment: () => number;
decrement: () => number;
reset: () => number;
};
function createCounter(init: number): ReturnObj {
let val = init;
return {
increment() {
return ++val;
},
decrement() {
return --val;
},
reset() {
return (val = init);
},
};
}
/**
* const counter = createCounter(5)
* counter.increment(); // 6
* counter.reset(); // 5
* counter.decrement(); // 4
*/
复杂度分析
| 指标 | 值 |
|---|---|
| 时间 | Depends on the final approach |
| 空间 | Depends on the final approach |
面试官常问的追问
外企场景- question_mark
Focus on correctly implementing the counter functions and their interactions with the initial state.
- question_mark
Ensure that each function operates independently and maintains the expected results.
- question_mark
Look for clear, concise, and efficient code that directly addresses the problem requirements.
常见陷阱
外企场景- error
Forgetting to preserve the initial state of the counter, leading to incorrect results when `reset` is called.
- error
Not handling edge cases, such as calling `reset` multiple times or performing `increment` and `decrement` in rapid succession.
- error
Overcomplicating the solution by introducing unnecessary variables or steps.
进阶变体
外企场景- arrow_right_alt
Modifying the counter object to support additional operations like `multiply` or `divide`.
- arrow_right_alt
Making the counter operations support negative increments or decrements.
- arrow_right_alt
Adding support for non-integer initial values (e.g., floating point numbers).