#371 Print Numbers by Recursion

最新推荐文章于 2023-08-03 02:13:00 发布

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最新推荐文章于 2023-08-03 02:13:00 发布

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分类专栏： lintcode 文章标签： lintcode recursion numbers

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本文介绍了一种使用递归算法以限定递归深度的方式打印从1到最大的N位数的方法。通过递归调用，每次输出特定位数的数字，并确保递归深度不超过N。给出的代码实现了这一目标，并提供了具体的实现细节。

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题目描述：

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].

Challenge

Do it in recursion, not for-loop.

题目思路：

既然要求N depth，那就是每次recursion都print i-digit的数字，比如第一次print 1~9，第二次print 10~99，etc。

Mycode（AC = 182ms）：

class Solution {
public:
    /**
     * @param n: An integer.
     * return : An array storing 1 to the largest number with n digits.
     */
    vector<int> numbersByRecursion(int n) {
        // write your code here
        vector<int> ans;
        numbersByRecursion(ans, 1, 10, 1, n);
        return ans;
    }
    
    void numbersByRecursion(vector<int>& ans, int start, int end, int level, int n) {
        if (level > n) {
            return;
        }
        
        // output numbers with 'level'-digits
        for (int i = start; i < end; i++) {
            ans.push_back(i);
        }
        
        numbersByRecursion(ans, end, end * 10, level + 1, n);
    }
};