351. Android Unlock Patterns

最新推荐文章于 2021-05-27 11:34:00 发布

原创最新推荐文章于 2021-05-27 11:34:00 发布 · 356 阅读

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#LeetCode #Dynamic Programming #Backtracking #Android Unlock Patte

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本文探讨了在Android 3x3锁屏模式下，如何计算由最少m个到最多n个键组成的有效解锁模式的数量。文章定义了有效模式的规则，并通过深度优先遍历的方法给出了解决方案。

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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:

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

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

深度优先遍历解题。建立vis数组存储当前数有没有被访问过，skip数组存储那两个数之间可以跳跃，之后就向下进行一层层的深度遍历。代码如下：

public class Solution {
    public int numberOfPatterns(int m, int n) {
        int[][] skip = new int[10][10];
        skip[1][3] = skip[3][1] = 2;
        skip[1][7] = skip[7][1] = 4;
        skip[7][9] = skip[9][7] = 8;
        skip[9][3] = skip[3][9] = 6;
        skip[1][9] = skip[9][1] = skip[7][3] = skip[3][7] = skip[2][8] = skip[8][2] = skip[4][6] = skip[6][4] =5;
        boolean[] vis = new boolean[10];
        int count = 0;
        for (int i = m; i <= n; i ++) {
            count += DFS(skip, vis, 1, i - 1) * 4;
            count += DFS(skip, vis, 4, i - 1) * 4;
            count += DFS(skip, vis, 5, i - 1);
        }
        return count;
    }
    
    private int DFS(int[][] skip, boolean[] vis, int curr, int remain) {
        if (remain < 0) {
            return 0;
        } else if (remain == 0) {
            return 1;
        }
        vis[curr] = true;
        int res = 0;
        for (int i = 1; i <= 9; i ++) {
            if (!vis[i] && (skip[curr][i] == 0 || vis[skip[curr][i]])) {
                res += DFS(skip, vis, i, remain - 1);
            }
        }
        vis[curr] = false;
        return res;
    }
}