Print Numbers by Recursion
最后更新于：2022-04-02 01:09:54
# 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 numbersByRecursion(int n) {
List result = new ArrayList();
if (n <= 0) {
return result;
}
helper(n, result);
return result;
}
private void helper(int n, List 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/)
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