Climbing Stairs

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);
        }
    }

}