LeetCode 题解工作台
时钟指针的夹角
给你两个数 hour 和 minutes 。请你返回在时钟上,由给定时间的时针和分针组成的较小角的角度(60 单位制)。 示例 1: 输入: hour = 12, minutes = 30 输出: 165 示例 2: 输入: hour = 3, minutes = 30 输出; 75 示例 3: 输…
1
题型
5
代码语言
3
相关题
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中等 · 数学·driven
答案摘要
时针每小时移动 30 度,每分钟移动 0.5 度。分针每分钟移动 6 度。如果指针之间的夹角大于 180 度,则取其与 360 度的差值,以确保获得最小的夹角。 时间复杂度 ,空间复杂度 。
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题目描述
给你两个数 hour 和 minutes 。请你返回在时钟上,由给定时间的时针和分针组成的较小角的角度(60 单位制)。
示例 1:

输入:hour = 12, minutes = 30 输出:165
示例 2:

输入:hour = 3, minutes = 30 输出;75
示例 3:

输入:hour = 3, minutes = 15 输出:7.5
示例 4:
输入:hour = 4, minutes = 50 输出:155
示例 5:
输入:hour = 12, minutes = 0 输出:0
提示:
1 <= hour <= 120 <= minutes <= 59- 与标准答案误差在
10^-5以内的结果都被视为正确结果。
解题思路
方法一:数学
时针每小时移动 30 度,每分钟移动 0.5 度。分针每分钟移动 6 度。如果指针之间的夹角大于 180 度,则取其与 360 度的差值,以确保获得最小的夹角。
时间复杂度 ,空间复杂度 。
class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
h = 30 * hour + 0.5 * minutes
m = 6 * minutes
diff = abs(h - m)
return min(diff, 360 - diff)
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
Pay attention to how the minute hand shifts the hour hand slightly as time progresses.
- question_mark
Candidates should show awareness of the fact that the angle must be adjusted for the smaller value, ensuring the correct result is found.
- question_mark
Look for clarity in the explanation of the mechanics of both hands and how their positions affect the angle.
常见陷阱
外企场景- error
Forgetting to account for the minute hand's effect on the hour hand.
- error
Not adjusting the angle when it exceeds 180 degrees to ensure the smaller angle is returned.
- error
Misunderstanding how to convert time into angles, leading to incorrect calculations.
进阶变体
外企场景- arrow_right_alt
Return the larger angle instead of the smaller one.
- arrow_right_alt
Calculate the angle between the hands at a different scale, like a 24-hour format clock.
- arrow_right_alt
Consider a case where the input time is provided in 24-hour format and adjust the formula accordingly.