G面经prepare: Android Phone Unlock Pattern

1 2 3
4 5 6
7 8 9
只有中间没有其他键的两个键才能相连，比如1可以连 2 4 5 6 8 但不能连 3 7 9
但是如果中间键被使用了，那就可以连，比如5已经被使用了，那1就可以连9
每个键只能用一次，给定一个长度L，求问有多少unique path with length L

Backtracking: 我的code不光可以知道数目，还可以打印所有Pattern

package AndroidUnlockPattern;
import java.util.*;

public class Solution {
    int res = 0;
    HashSet<Integer> set = new HashSet<Integer>();
    static ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
    ArrayList<Integer> path = new ArrayList<Integer>();
    
    public int calculate(int L) {
        for (int i=1; i<=9; i++) {
            set.add(i);
        }
        HashSet<Integer> visited = new HashSet<Integer>();
        helper(0, 0, L, visited);
        return res;
    }
    
    public void helper(int cur, int pos, int L, HashSet<Integer> visited) {
        if (pos == L) {
            res++;
            result.add(new ArrayList<Integer>(path));
            return;
        }
        for (int elem : set) {
            if (visited.contains(elem)) continue;
            if (cur == 1) {
                if (elem==3 && !visited.contains(2)) continue;
                if (elem==7 && !visited.contains(4)) continue;
                if (elem==9 && !visited.contains(5)) continue;
            }
            else if (cur == 2) {
                if (elem==8 && !visited.contains(5)) continue;
            }
            else if (cur == 3) {
                if (elem==1 && !visited.contains(2)) continue;
                if (elem==7 && !visited.contains(5)) continue;
                if (elem==9 && !visited.contains(6)) continue;
            }
            else if (cur == 4) {
                if (elem == 6 && !visited.contains(5)) continue;
            }
            else if (cur == 6) {
                if (elem == 4 && !visited.contains(5)) continue;
            }
            else if (cur == 7) {
                if (elem==1 && !visited.contains(4)) continue;
                if (elem==3 && !visited.contains(5)) continue;
                if (elem==9 && !visited.contains(8)) continue;
            }
            else if (cur == 8) {
                if (elem==2 && !visited.contains(5)) continue;
            }
            else if (cur == 9) {
                if (elem==1 && !visited.contains(5)) continue;
                if (elem==3 && !visited.contains(6)) continue;
                if (elem==7 && !visited.contains(8)) continue;
            }
            visited.add(elem);
            path.add(elem);
            helper(elem, pos+1, L, visited);
            visited.remove(elem);
            path.remove(path.size()-1);
        }
    }
    

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution sol = new Solution();
        int res = sol.calculate(3);
        System.out.println(res);
        for (ArrayList<Integer> each : result) {
            System.out.println(each);
        }
    }

}