算法

(模拟) $O(1)$

一种思路是分别计算出时针与0时刻的夹角，以及分针与0时刻的夹角，然后对二者求一个绝对差，而由于可能会是钝角，所以得取min(dist, 360 - dist)

另一种思路是十分和分针的夹角是5.4 * (hour * 60 + minutes) % 360，然后取小于等于$180$度的角

Java 代码

class Solution {
    public double angleClock(int hour, int minutes) {
        // 360 / 12 = 30
        // 360 / 60 = 6
        double hourHand = (hour + minutes / 60.0) * 30;
        double minHand = minutes * 6;
        double dist = Math.abs(hourHand - minHand);
        return Math.min(dist, 360 - dist);
    }
}

Python 代码

class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        angle = 5.5 * (hour * 60 + minutes) % 360
        return angle if angle <= 180 else 360 - angle