Print Numbers by Recursion

Print numbers from 1 to the largest number with N digits by recursion.

Notice

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?

Have you met this question in a real interview?

Yes

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].

分析：

我们可以按“层”打印，什么意思呢？

第一层 1 - 9  = 1 to (pow(10, 1) - pow(10, 0))

第二层： 10 - 99 = 10 to (pow(10, 2) - pow(10, 1))

第三层： 100 - 999 = 100 to (pow(10, 3) - pow(10, 2))

...

看到规律了吧。

 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> list = new ArrayList<Integer>();
         if (n <= ) return list;
         // start[0] refers to the start number for the current levevl.
         // start[1] refers to the exponent.
         int[] start = {, };
         helper(n, list, start);
         return list;
     }
 
     public void helper(int n, List<Integer> list, int[] start) {
         if (n == ) return;
         for (int i = ; i <= (int)(Math.pow(, start[]) - Math.pow(, start[] - )); i++) {
             list.add(start[]++);
         }
         start[]++;
         helper(n - , list, start);
     }
 }