[LintCode] 用递归打印数字 Print Numbers by Recursion

用递归的方法找到从1到最大的N位整数。

你能够用深度最多只有 N 层的递归么？

样例
给出 N = 1, 返回[1,2,3,4,5,6,7,8,9].
给出 N = 2, 返回[1,2,3,4,5,6,7,8,9,10,11,…,99].

Print numbers from 1 to the largest number with N digits by recursion.
Can you recursive with at most N depth?

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].

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) {
        if(n <= 0) {
            return new ArrayList<Integer>();
        }else if(n == 1) {
            List<Integer> result = new ArrayList<Integer>();
            for(int i = 1; i <= 9; i++) {
                result.add(i);
            }
            return result;
        }else {
            List<Integer> list = numbersByRecursion(n-1);
            List<Integer> result = new ArrayList<Integer>(list);
            for(int i = 1; i <= 9; i++) {
                int topdigit = (int)Math.pow(10, n-1) * i;
                result.add(topdigit);
                for(int j = 0; j < list.size(); j++) {
                    result.add(topdigit + list.get(j));
                }
            }
            return result;
        }
    }
}