1344. 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 <= 5910^-5 以内的结果都被视为正确结果。原站题解
java 解法, 执行用时: 0 ms, 内存消耗: 38.7 MB, 提交时间: 2022-12-05 15:34:40
class Solution {
public double angleClock(int hour, int minutes) {
int oneMinAngle = 6;
int oneHourAngle = 30;
double minutesAngle = oneMinAngle * minutes;
double hourAngle = (hour % 12 + minutes / 60.0) * oneHourAngle;
double diff = Math.abs(hourAngle - minutesAngle);
return Math.min(diff, 360 - diff);
}
}
golang 解法, 执行用时: 0 ms, 内存消耗: 1.9 MB, 提交时间: 2022-12-05 15:34:26
func angleClock(hour int, minutes int) float64 {
var oneMinAngle, oneHourAngle, minutesAngle, hourAngle, diff float64;
oneMinAngle = 6;
oneHourAngle = 30;
minutesAngle = oneMinAngle * float64(minutes);
hourAngle = (float64(hour % 12) + float64(minutes) / 60.0) * oneHourAngle;
diff = math.Abs(hourAngle - minutesAngle);
return math.Min(diff, 360 - diff);
}
python3 解法, 执行用时: 36 ms, 内存消耗: 14.9 MB, 提交时间: 2022-12-05 15:34:13
class Solution:
def angleClock(self, hour: int, minutes: int) -> float:
one_min_angle = 6
one_hour_angle = 30
minutes_angle = one_min_angle * minutes
hour_angle = (hour % 12 + minutes / 60) * one_hour_angle
diff = abs(hour_angle - minutes_angle)
return min(diff, 360 - diff)