README.md

comments	difficulty	edit_url	rating	source	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/1300-1399/1344.Angle%20Between%20Hands%20of%20a%20Clock/README.md	1324	第 19 场双周赛 Q3	数学

1344. 时钟指针的夹角

English Version

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 <= 12
• 0 <= 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。

解法

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 $O(1)$，空间复杂度 $O(1)$。

Python3

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)

Java

class Solution {
    public double angleClock(int hour, int minutes) {
        double h = 30 * hour + 0.5 * minutes;
        double m = 6 * minutes;
        double diff = Math.abs(h - m);
        return Math.min(diff, 360 - diff);
    }
}

C++

class Solution {
public:
    double angleClock(int hour, int minutes) {
        double h = 30 * hour + 0.5 * minutes;
        double m = 6 * minutes;
        double diff = abs(h - m);
        return min(diff, 360 - diff);
    }
};

Go

func angleClock(hour int, minutes int) float64 {
	h := 30*float64(hour) + 0.5*float64(minutes)
	m := 6 * float64(minutes)
	diff := math.Abs(h - m)
	return math.Min(diff, 360-diff)
}

TypeScript

function angleClock(hour: number, minutes: number): number {
    const h = 30 * hour + 0.5 * minutes;
    const m = 6 * minutes;
    const diff = Math.abs(h - m);
    return Math.min(diff, 360 - diff);
}