[LeetCode July Challange]Day14-Angle Between Hands of a Clock

Given two numbers, hour and minutes. Return the smaller angle (in degrees) formed between the hour and the minute hand.

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

Example 4:

Input: hour = 4, minutes = 50
Output: 155

Example 5:

Input: hour = 12, minutes = 0
Output: 0

Constraints:

• 1 <= hour <= 12
• 0 <= minutes <= 59
• Answers within 10^-5 of the actual value will be accepted as correct.

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solution

time complexity : O(1)
space complexity : O(1)

class Solution {
public:
    double angleClock(int hour, int minutes) {
        // total angle = units * angle per unit
        double minAngle = minutes * 360/60;
        double hourAngle = (hour+minutes/60.0) * 360/12;
        
        double res = abs(minAngle-hourAngle);
        return min(res, 360-res);
    }
};

計算每一分鐘，分針走的角度。
計算每一分鐘、小時，時針走的角度。
有了每單位走的角度，就可用總量來算出角度。
相減即是夾角。
因算出來的角度有可能會是大角，所以與小角取最小值。