Description
Given two numbers, hour and minutes, return the smaller angle (in degrees) formed between the hour and the minute hand.
Answers within 10-5 of the actual value will be accepted as correct.
Example 1:
Input: hour = 12, minutes = 30
Output: 165
Example 2:
Input: hour = 3, minutes = 30
Output: 75
Example 3:
Input: hour = 3, minutes = 15
Output: 7.5
Constraints:
1 <= hour <= 120 <= minutes <= 59
Solutions
Solution 1
PythonJavaC++GoTypeScript
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)(code-box)
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);
}
}(code-box)
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);
}
};(code-box)
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)
}(code-box)
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);
}(code-box)
