Climbing Stairs - Print Path

stair climbing， print out all of possible solutions of the methods to climb a stars, you are allowed climb one or two steps for each time; what is time/space complexity? （use recursion）

这道题难是难在这个ArrayList<String> res是用在argument还是返回值，纠结了好久

Recursion 解法：

 package fib;

 import java.util.ArrayList;

 public class climbingstairs {

     public ArrayList<String> climb (int n) {
         if (n <= 0) return null;
         ArrayList<String> res = new ArrayList<String>();
         if (n == 1) {
             res.add("1");
             return res;
         }
         if (n == 2) {
             res.add("2");
             res.add("12");
             return res;
         }
         ArrayList<String> former2 =  climb(n-2);
         for (String item : former2) {
             res.add(item+Integer.toString(n));
         }
         ArrayList<String> former1 = climb(n-1);
         for (String item : former1) {
             res.add(item+Integer.toString(n));
         }
         return res;
     }

     public static void main(String[] args) {
         climbingstairs obj = new climbingstairs();
         ArrayList<String> res = obj.climb(6);
         for (String item : res) {
             System.out.println(item);
         }
     }

 }

Sample input : 6

Sample Output:

246
1246
1346
2346
12346
1356
2356
12356
2456
12456
13456
23456
123456

follow up: could you change the algorithm to save space?

这就想到DP，用ArrayList<ArrayList<String>>

 import java.util.ArrayList;

 public class climbingstairs {

     public ArrayList<String> climb (int n) {
         if (n <= 0) return null;
         ArrayList<ArrayList<String>> results = new ArrayList<ArrayList<String>>();
         for (int i=1; i<=n; i++) {
             results.add(new ArrayList<String>());
         }
         if (n >= 1) {
             results.get(0).add("1");
         }
         if (n >= 2) {
             results.get(1).add("2");
             results.get(1).add("12");
         }

         for (int i=3; i<=n; i++) {
             ArrayList<String> step = results.get(i-1);
             ArrayList<String> former2 = results.get(i-3);
             for (String item : former2) {
                 step.add(item+Integer.toString(i));
             }
             ArrayList<String> former1 = results.get(i-2);
             for (String item : former1) {
                 step.add(item+Integer.toString(i));
             }
         }
         return results.get(n-1);
     }

     public static void main(String[] args) {
         climbingstairs obj = new climbingstairs();
         ArrayList<String> res = obj.climb(5);
         for (String item : res) {
             System.out.println(item);
         }
     }

 }