Print Numbers by Recursion详解编程语言

Print numbers from 1 to the largest number with N digits by recursion. 
 
Example 
Given N = 1, return [1,2,3,4,5,6,7,8,9]. 
 
Given N = 2, return [1,2,3,4,5,6,7,8,9,10,11,12,...,99]. 
 
Note 
It's pretty easy to do recursion like: 
 
recursion(i) { 
    if i > largest number: 
        return 
    results.add(i) 
    recursion(i + 1) 
} 
however this cost a lot of recursion memory as the recursion depth maybe very large. 
Can you do it in another way to recursive with at most N depth? 
 
Challenge 
Do it in recursion, not for-loop.

public class Solution { 
    /** 
     * @param n: An integer. 
     * return : An array storing 1 to the largest number with n digits. 
     */ 
    public List<Integer> numbersByRecursion(int n) { 
        List<Integer> result = new ArrayList<Integer>(); 
        if (n <= 0) { 
            return result; 
        } 
        helper(n, result); 
        return result; 
    } 
 
    private void helper(int n, List<Integer> ret) { 
        if (n == 0) return; 
        helper(n - 1, ret); 
        // current base such as 10, 20, 30... 
        int base = (int)Math.pow(10, n - 1); 
        // get List size before for loop 
        int size = ret.size(); 
        for (int i = 1; i < 10; i++) { 
            // add 10, 100, 1000... 
            ret.add(i * base); 
            for (int j = 0; j < size; j++) { 
                // add 11, 12, 13... 
                ret.add(ret.get(j) + base * i); 
            } 
        } 
    } 
}