[Swift]LeetCode858. 鏡面反射 | Mirror Reflection

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There is a special square room with mirrors on each of the four walls.  Except for the southwest corner, there are receptors on each of the remaining corners, numbered 0, 1, and 2.

The square room has walls of length p, and a laser ray from the southwest corner first meets the east wall at a distance q from the 0th receptor.

Return the number of the receptor that the ray meets first.  (It is guaranteed that the ray will meet a receptor eventually.) 

Example 1:

Input: p = 2, q = 1 Output: 2 Explanation: The ray meets receptor 2 the first time it gets reflected back to the left wall.

Note:

1. 1 <= p <= 1000
2. 0 <= q <= p

---

有一個特殊的正方形房間，每面牆上都有一面鏡子。除西南角之外，每一個角落都放有一個接受器，編號爲 0， 1，以及 2。

正方形房間的牆壁長度爲 p，一束激光從西南角射出，首先會與東牆相遇，入射點到接收器 0 的距離爲 q 。

返回光線最早遇到的接收器的編號（保證光線最終會遇到一個接收器）。 

示例：

輸入： p = 2, q = 1
輸出： 2
解釋： 這條光線在第一次被反射回左邊的牆時就遇到了接收器 2 。

 

提示：

1. 1 <= p <= 1000
2. 0 <= q <= p

---

Runtime: 4 ms

Memory Usage: 18.5 MB

 1 class Solution {
 2     func mirrorReflection(_ p: Int, _ q: Int) -> Int {
 3         var p = p
 4         var q = q
 5         while (p % 2 == 0 && q % 2 == 0)
 6         {
 7             p /= 2
 8             q /= 2
 9         }
10         if p % 2 == 0
11         {
12             return 2
13         }
14         else if q % 2 == 0
15         {
16             return 0
17         }
18         else
19         {
20              return 1
21         }
22     }
23 }

---

4ms

 1 final class Solution {
 2     func mirrorReflection(_ p: Int, _ q: Int) -> Int {
 3         var curQ = 0
 4         var count = 0
 5         var opposite = false
 6         while curQ != p {
 7             curQ &+= q
 8             count &+= 1
 9             if curQ > p {
10                 curQ &-= p
11                 opposite = !opposite
12             }
13         }
14         if opposite { return 0 }
15         return count % 2 == 1 ? 1 : 2
16     }
17 }

---

8ms

 1 class Solution {
 2     func mirrorReflection(_ p: Int, _ q: Int) -> Int {
 3         var n=1
 4         while Double(n*q)/Double(p)-Double(n*q/p) != 0 {
 5             n+=1
 6         }
 7         if Double(n*q/p)/Double(2) - Double(n*q/(2*p))==0 {
 8             return Double(n)/Double(2) - Double(n/2)==0 ? -1:0
 9         }else{
10             return Double(n)/Double(2) - Double(n/2)==0 ? 2:1
11         }
12     }
13 }