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:
- Each pattern must connect at least m keys and at most n keys.
- All the keys must be distinct.
- 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.
- 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:
Input: m = 1, n = 1 Output: 9
Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.
這道題乍一看題目這么長以為是一個設計題，其實不是，這道題還是比較有意思的，起碼跟實際結合的比較緊密。這道題說的是安卓機子的解鎖方法，有9個數字鍵，如果密碼的長度范圍在 [m, n] 之間，問所有的解鎖模式共有多少種，注意題目中給出的一些非法的滑動模式。那么先來看一下哪些是非法的，首先1不能直接到3，必須經過2，同理的有4到6，7到9，1到7，2到8，3到9，還有就是對角線必須經過5，例如1到9，3到7等。建立一個二維數組 jumps，用來記錄兩個數字鍵之間是否有中間鍵，然后再用一個一位數組 visited 來記錄某個鍵是否被訪問過，然后用遞歸來解，先對1調用遞歸函數，在遞歸函數中，遍歷1到9每個數字 next，然后找他們之間是否有 jump 數字，如果 next 沒被訪問過，並且 jump 為0，或者 jump 被訪問過，對 next 調用遞歸函數。數字1的模式個數算出來后，由於 1,3,7,9 是對稱的，所以乘4即可，然后再對數字2調用遞歸函數，2,4,6,9 也是對稱的，再乘4，最后單獨對5調用一次，然后把所有的加起來就是最終結果了，參見代碼如下：
解法一：
class Solution { public: int numberOfPatterns(int m, int n) { int res = 0; vector<bool> visited(10, false); vector<vector<int>> jumps(10, vector<int>(10, 0)); jumps[1][3] = jumps[3][1] = 2; jumps[4][6] = jumps[6][4] = 5; jumps[7][9] = jumps[9][7] = 8; jumps[1][7] = jumps[7][1] = 4; jumps[2][8] = jumps[8][2] = 5; jumps[3][9] = jumps[9][3] = 6; jumps[1][9] = jumps[9][1] = jumps[3][7] = jumps[7][3] = 5; res += helper(1, 1, m, n, jumps, visited, 0) * 4; res += helper(2, 1, m, n, jumps, visited, 0) * 4; res += helper(5, 1, m, n, jumps, visited, 0); return res; } int helper(int num, int len, int m, int n, vector<vector<int>>& jumps, vector<bool>& visited, int res) { if (len >= m) ++res; ++len; if (len > n) return res; visited[num] = true; for (int next = 1; next <= 9; ++next) { int jump = jumps[num][next]; if (!visited[next] && (jump == 0 || visited[jump])) { res = helper(next, len, m, n, jumps, visited, res); } } visited[num] = false; return res; } };
下面這種方法很簡潔，但是不容易理解，講解請看這個帖子。其中 used 是一個9位的 mask，每位對應一個數字，如果為1表示存在，0表示不存在，(i1, j1) 是之前的位置，(i, j) 是當前的位置，所以滑動是從 (i1, j1) 到 (i, j)，中間點為 ((i1+i)/2, (j1+j)/2)，這里的I和J分別為 i1+i 和 j1+j，還沒有除以2，所以I和J都是整數。如果 I%2 或者 J%2 不為0，說明中間點的坐標不是整數，即中間點不存在，如果中間點存在，如果中間點被使用了，則這條線也是成立的，可以調用遞歸，參見代碼如下：
解法二：
class Solution { public: int numberOfPatterns(int m, int n) { return count(m, n, 0, 1, 1); } int count(int m, int n, int used, int i1, int j1) { int res = m <= 0; if (!n) return 1; for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { int I = i1 + i, J = j1 + j, used2 = used | (1 << (i * 3 + j)); if (used2 > used && (I % 2 || J % 2 || used2 & (1 << (I / 2 * 3 + J / 2)))) { res += count(m - 1, n - 1, used2, i, j); } } } return res; } };
Github 同步地址：
https://github.com/grandyang/leetcode/issues/351
參考資料：
https://leetcode.com/problems/android-unlock-patterns/