1344. Angle Between Hands of a Clock

題目 / Problem

中文： 給定兩個數字 hour（時）和 minutes（分），回傳時針與分針之間形成的較小夾角（以度為單位）。只要答案與正確值的誤差在 10⁻⁵ 以內就算正確。

English: Given two integers hour and minutes, return the smaller angle (in degrees) formed between the hour hand and the minute hand of a clock. Any answer within 10⁻⁵ of the actual value is accepted.

Constraints / 約束： - 1 <= hour <= 12 - 0 <= minutes <= 59

Worked example / 範例： hour = 12, minutes = 30 → 輸出 165。在 12:30 時，分針指向 6（180°），時針從 12 走到了 12 與 1 之間（已走過半小時，即 15°），兩者相差 180 - 15 = 165 度。

名詞解釋 / Glossary

• 度 / degree（°）：圓周分成 360 等份，每一份就是 1 度。整個鐘面是 360°。 A full circle is 360 degrees; the whole clock face spans 360°.
• 絕對值 / absolute value：一個數去掉正負號後的大小，例如 |-5| = 5。C 裡用 fabs（浮點數絕對值）來計算。The magnitude of a number ignoring its sign; we use fabs for floating-point numbers.
• 取餘數 / modulo (%)：求除法剩下的餘數。13 % 12 = 1。這裡用來把 12 點當成 0 點處理。Gives the remainder of a division; used to treat 12 o'clock as 0.
• 浮點數 / floating-point (double)：可以表示小數的數字型別，因為角度可能是 7.5 這種非整數。A number type that stores decimals, needed because angles like 7.5 aren't whole numbers.

思路

中文： 這題的核心是「角度」。一個完整的鐘面是 360 度，分成 12 個小時刻度，所以每小時佔 30 度（360 ÷ 12）；分針則把整圈分成 60 格，所以每分鐘佔 6 度（360 ÷ 60）。

最直接的想法：分別算出分針和時針各自離 12 點的角度，再相減取絕對值。分針很簡單，minutes × 6。時針的陷阱（也是 hint 提示的地方）在於：時針不是固定停在整點，它會隨著分鐘慢慢往前移動。例如 12:30，時針已經走到 12 和 1 的中間。時針每小時走 30 度，那麼每一分鐘它額外走 30 / 60 = 0.5 度。所以時針的角度是 hour × 30 + minutes × 0.5。

要注意 hour = 12 時應當視為 0（指向正上方），用 hour % 12 處理。算出兩個角度後相減取絕對值得到 diff。但這個 diff 可能大於 180（例如算出 350 度），而題目要「較小」的夾角，這時真正的較小角是 360 - diff。所以最後回傳 min(diff, 360 - diff)。整個過程只是幾個算式，沒有迴圈，是 O(1)。

English: The whole idea is converting clock positions into degrees. A full clock is 360°. Since there are 12 hour marks, each hour is 30° (360 ÷ 12). Since there are 60 minutes around the dial, each minute is 6° (360 ÷ 60).

The straightforward plan: compute each hand's angle measured from 12, then take the absolute difference. The minute hand is easy — minutes × 6. The trap (the part the hints warn about) is the hour hand: it doesn't sit still on the hour, it creeps forward as minutes pass. At 12:30 the hour hand is halfway between 12 and 1. Since it moves 30° per hour, it moves an extra 30 / 60 = 0.5° per minute. So the hour hand's angle is hour × 30 + minutes × 0.5.

Treat hour = 12 as 0 (pointing straight up) using hour % 12. After getting both angles, subtract and take the absolute value to get diff. But diff might exceed 180° (say it comes out as 350°), and the problem wants the smaller angle — in that case the smaller side is 360 - diff. So we return min(diff, 360 - diff). It's all arithmetic, no loops, so it runs in O(1).

逐步走查 / Walkthrough

Example input / 範例輸入：hour = 12, minutes = 30

時針只走了 15 度（因為過了半小時），分針指向正下方 180 度，相差 165 度，且 165 < 195，所以答案是 165。

Solution — C

// 演算法：把時針、分針各自轉成「離 12 點的角度」，相減取絕對值。
// Algorithm: convert each hand to its angle from 12 o'clock, then take the
// absolute difference. The hour hand also drifts 0.5° for every minute.
// 最後若夾角大於 180，較小角是 360 減去它 / If the gap exceeds 180°, the
// smaller angle is 360 minus it. All O(1) arithmetic.

#include <math.h>   // 提供 fabs（浮點絕對值）/ provides fabs (float absolute value)

double angleClock(int hour, int minutes) {
    // 時針角度：hour%12 把 12 點當成 0；每小時 30 度；每分鐘額外 0.5 度
    // Hour-hand angle: hour%12 maps 12→0; 30° per hour; plus 0.5° per minute
    double hourAngle = (hour % 12) * 30.0 + minutes * 0.5;

    // 分針角度：每分鐘 6 度（360 度 ÷ 60 分）
    // Minute-hand angle: 6° per minute (360° ÷ 60 minutes)
    double minuteAngle = minutes * 6.0;

    // 兩角相差，取絕對值（fabs 去掉正負號）/ absolute difference of the two angles
    double diff = fabs(hourAngle - minuteAngle);

    // 夾角不能超過 180；若超過就走另一邊 360-diff，取較小者
    // The angle can't exceed 180°; if it does, the other side (360-diff) is smaller
    return diff < 360.0 - diff ? diff : 360.0 - diff;
}

Solution — C++

// 演算法與 C 版完全相同：時針角度 = hour*30 + minutes*0.5，分針角度 = minutes*6，
// Algorithm identical to the C version: hour angle = hour*30 + minutes*0.5,
// minute angle = minutes*6, take |difference|, then the smaller of it and 360-it.
// 取兩角差的絕對值，再回傳它與 360 減它之中較小的一個。O(1)。

#include <cmath>      // std::fabs / std::min 用到 / for fabs
#include <algorithm>  // std::min 提供較小值 / provides std::min

class Solution {
public:
    double angleClock(int hour, int minutes) {
        // hour % 12 讓 12 點對齊 0 度；每小時 30 度，每分鐘再加 0.5 度
        // hour % 12 aligns 12 o'clock to 0°; 30° per hour, plus 0.5° per minute
        double hourAngle = (hour % 12) * 30.0 + minutes * 0.5;

        // 分針：每分鐘 6 度 / minute hand: 6° per minute
        double minuteAngle = minutes * 6.0;

        // std::fabs 取浮點絕對值，得到兩針之間的角差
        // std::fabs gives the floating-point absolute value (the raw gap)
        double diff = std::fabs(hourAngle - minuteAngle);

        // std::min 回傳兩者中較小的；夾角與其補角(360-diff)取小
        // std::min returns the smaller of the gap and its complement (360 - gap)
        return std::min(diff, 360.0 - diff);
    }
};

複雜度 / Complexity

• Time: O(1) — 只是固定幾個乘法、減法和一次比較，與輸入大小無關，沒有任何迴圈。Just a fixed handful of multiplications, a subtraction, and one comparison — no loops, independent of input size.
• Space: O(1) — 只用了 hourAngle、minuteAngle、diff 三個變數，不隨輸入增長。Only three scalar variables are used; nothing grows with the input.

Pitfalls & Edge Cases

• 忘記時針會移動 / Forgetting the hour hand drifts：最常見的錯誤是把時針角度寫成 hour * 30，沒加上 minutes * 0.5。這樣 3:30 會算錯。The most common bug is using hour * 30 alone and dropping the minutes * 0.5 term — that breaks cases like 3:30.
• 12 點沒歸零 / Not wrapping 12 to 0：hour = 12 若直接乘 30 會得到 360 度而非 0 度。用 hour % 12 修正。Without hour % 12, twelve o'clock becomes 360° instead of 0°.
• 沒取較小夾角 / Returning the wrong side：差值可能 > 180（例如 210），題目要較小角，必須跟 360 - diff 比較取小。The raw difference can exceed 180°; you must compare against 360 - diff and keep the smaller.
• 用整數導致小數遺失 / Integer division losing decimals：角度可能是 7.5，務必用 double 而非 int，且寫 0.5、6.0 等浮點常數。Angles like 7.5 require double; using int would truncate the result.