LeetCode 351. Android Unlock Patterns（安卓解锁）

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本文介绍了一个基于深度优先搜索算法的问题解决方法，该方法用于计算3x3键盘锁屏上从m到n个按键的所有有效解锁模式数量。文章详细阐述了有效解锁模式的规则，并给出了具体的实现代码。

原题网址：https://leetcode.com/problems/android-unlock-patterns/

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.

方法：深度优先搜索。

public class Solution {
    private int patterns;
    private boolean valid(boolean[] keypad, int from, int to) {
        if (from==to) return false;
        int i=Math.min(from, to), j=Math.max(from,to);
        if ((i==1 && j==9) || (i==3 && j==7)) return keypad[5] && !keypad[to];
        if ((i==1 || i==4 || i==7) && i+2==j) return keypad[i+1] && !keypad[to];
        if (i<=3 && i+6==j) return keypad[i+3] && !keypad[to];
        return !keypad[to];
    }
    private void find(boolean[] keypad, int from, int step, int m, int n) {
        if (step == n) {
            patterns ++;
            return;
        }
        if (step >= m) patterns ++;
        for(int i=1; i<=9; i++) {
            if (valid(keypad, from, i)) {
                keypad[i] = true;
                find(keypad, i, step+1, m, n);
                keypad[i] = false;
            }
        }
    }
    public int numberOfPatterns(int m, int n) {
        boolean[] keypad = new boolean[10];
        for(int i=1; i<=9; i++) {
            keypad[i] = true;
            find(keypad, i, 1, m, n);
            keypad[i] = false;
        }
        return patterns;
    }
}