# Print Numbers by Recursion ### Source - lintcode: [(371) Print Numbers by Recursion](http://www.lintcode.com/en/problem/print-numbers-by-recursion/) ~~~ 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. ~~~ ### 题解 从小至大打印 N 位的数列，正如题目中所提供的 `recursion(i)`, 解法简单粗暴，但问题在于 N 稍微大一点时栈就溢出了，因为递归深度太深了。能联想到的方法大概有两种，一种是用排列组合的思想去解释，把0~9当成十个不同的数(字符串表示)，塞到 N 个坑位中，这个用 [DFS](# "Depth-First Search, 深度优先搜索") 来解应该是可行的；另一个则是使用数学方法，依次递归递推，比如 N=2 可由 N=1递归而来，具体方法则是乘10进位加法。题中明确要求递归深度最大不超过 N, 故 [DFS](# "Depth-First Search, 深度优先搜索") 方法比较危险。 ### Java ~~~ 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); } } } } ~~~ ### 源码分析 递归步的截止条件`n == 0`, 由于需要根据之前 N-1 位的数字递推，`base` 每次递归一层都需要乘10，`size`需要在`for`循环之前就确定。 ### 复杂度分析 添加 10n10^n10n 个元素，时间复杂度 O(10n)O(10^n)O(10n), 空间复杂度 O(1)O(1)O(1). ### Reference - [Lintcode: Print Numbers by Recursion | codesolutiony](https://codesolutiony.wordpress.com/2015/05/21/lintcode-print-numbers-by-recursion/)