[Swift]LeetCode351. 安卓解鎖模式 $ Android Unlock Patterns

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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Given an Android 3x3 key lock screen and two integers m and n, where 1 ≤ m ≤ n ≤ 9, count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys.

Rules for a valid pattern:

1. Each pattern must connect at least m keys and at most n keys.
2. All the keys must be distinct.
3. If the line connecting two consecutive keys in the pattern passes through any other keys, the other keys must have previously selected in the pattern. No jumps through non selected key is allowed.
4. The order of keys used matters.

Explanation:

| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |

Invalid move: 4 - 1 - 3 - 6 
Line 1 - 3 passes through key 2 which had not been selected in the pattern.

Invalid move: 4 - 1 - 9 - 2
Line 1 - 9 passes through key 5 which had not been selected in the pattern.

Valid move: 2 - 4 - 1 - 3 - 6
Line 1 - 3 is valid because it passes through key 2, which had been selected in the pattern

Valid move: 6 - 5 - 4 - 1 - 9 - 2
Line 1 - 9 is valid because it passes through key 5, which had been selected in the pattern.

Example:
Given m = 1, n = 1, return 9.

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

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給定一個android 3x3密鑰鎖屏和兩個整數m和n，其中1≤m≤n≤9，計算android鎖屏的解鎖模式總數，該解鎖模式由最小M個密鑰和最大N個密鑰組成。 

有效模式的規則：

1.  每一個模式必須鏈接至少M個鍵和最多N個鍵。
2. 全部鍵必須是不一樣的。
3. 若是鏈接模式中兩個連續鍵的線經過任何其餘鍵，則其餘鍵必須事先在模式中選擇。不容許跳過未選定的鍵。
4. 鑰匙的使用順序很重要。

說明：

| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |

無效移動：4-1-3-6

第1-3行經過模式中未選擇的鍵2。

無效移動：4-1-9-2

第1-9行經過模式中未選擇的鍵5。

有效移動：2-4-1-3-6

第1-3行有效，由於它經過模式中選擇的鍵2

有效移動：6-5-4-1-9-2

第1-9行有效，由於它經過模式中選擇的鍵5。

例子：

給定m=1，n=1，返回9。

信用：

特別感謝@elmirap添加此問題並建立全部測試用例。

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Solution:

 1 class Solution {
 2     func numberOfPatterns(_ m:Int,_ n:Int) -> Int
 3     {
 4         return count(m, n, 0, 1, 1)
 5     }
 6     
 7     func count(_ m:Int,_ n:Int,_ used:Int,_ i1:Int,_ j1:Int) -> Int
 8     {
 9         var res:Int = m <= 0 ? 1 : 0
10         if n == 0 {return 1}
11         for i in 0..<3
12         {
13             for j in 0..<3
14             {
15                 var I:Int = i1 + i
16                 var J:Int = j1 + j
17                 var used2:Int = used | 1 << (i * 3 + j)
18                 let num1:Int = ((I % 2 == 0) || (J % 2 == 0) || (used2 == 0)) ? 1 : 0
19                 let num2:Int = 1 << (I / 2 * 3 + J / 2)
20                 if used2 > used && (num1 & num2) == 0
21                 {
22                     res += count(m - 1, n - 1, used2, i, j)
23                 }
24             }
25         }
26         return res
27     }
28 }

點擊：Playground測試

1 var sol = Solution()
2 print(sol.numberOfPatterns(1,1))
3 //Print 9