LeetCode解题报告（101）-- 1344. Angle Between Hands of a Clock

Problem

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

Example 1:

1 2	Input: hour = 12, minutes = 30 Output: 165

Example 2:

1 2	Input: hour = 3, minutes = 30 Output: 75

Example 3:

1 2	Input: hour = 3, minutes = 15 Output: 7.5

Example 4:

1 2	Input: hour = 4, minutes = 50 Output: 155

Example 5:

1 2	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.

Analysis

这道题目是关于时间角度计算的一道简单数学题，要求的时候分针和时针的角度差。首先一圈是360度，对分针来说，有60分钟，所以一分钟就是6度；对时针来说，有12小时，所以一小时就是30度。

当然这里还需要注意的点是当分针有偏移的时候，时针也会有相对应的偏移，所以在计算时针的角度的时候需要把这个偏移算上。

Solution

分别计算时针和分针的偏移，然后取相减后的绝对值，最后用计算出较小的角度（min(diff, 360 - diff)）。

Code

class Solution {
public:
    double angleClock(int hour, int minutes) {
        double minAngle = minutes * 6;
        double hourAngle = (hour % 12) * 30 + minutes * 30 / 60.0;
        
        // cout << hourAngle << " " << minAngle << endl;
        double diff = fabs(hourAngle - minAngle);
        // cout << diff << endl;
        return min(diff, 360 - diff);
    }
};

Summary

这道题目的难度不太大，主要就是一个比较简单的数学问题如何转化为代码，写的过程要注意除法的小数和角度。希望这篇博客能够帮助到您，感谢您的支持，欢迎转发、分享、评论，谢谢！