【數學】C074_LC_時鐘指針的夾角（補角）

一、Problem

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

二、Solution

方法一：求補角

class Solution {
    public double angleClock(int h, int m) {
        double h2a = 360/12, m2a = h2a/60;	
        double h_angle = h2a*h + m2a*m;		//小時的貢獻
        double m_angle = m * 360/60;		//分鐘的貢獻
        double now = Math.abs(h_angle - 0 - m_angle - 0);
        double oth = 360 - now;				//補角
        return Math.min(now, oth);
    }
}

複雜度分析

• 時間複雜度：$O(1)$，
• 空間複雜度：$O(1)$，