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);
         }
     }

 }