用递归打印数字

371. 用递归打印数字

用递归的方法找到从1到最大的N位整数。

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].

挑战

用递归完成，而非循环的方式。

注意事项

用下面这种方式去递归其实很容易：

recursion(i) {
    if i > largest number:
        return
    results.add(i)
    recursion(i + 1)
}

但是这种方式会耗费很多的递归空间，导致堆栈溢出。你能够用其他的方式来递归使得递归的深度最多只有 N 层么？

class Solution:
    """
    @param n: An integer
    @return: An array storing 1 to the largest number with n digits.
    """
    def numbersByRecursion(self, n):
        # write your code here
        if n<1:
            return []
        largest=pow(10,n)
        res=[]
        for i in range(1,largest):
            res.append(i)
        return res
        

########################################################
class Solution {
public:
    /**
     * @param n: An integer
     * @return: An array storing 1 to the largest number with n digits.
     */
    vector numbersByRecursion(int n) {
        // write your code here
        vector res;
        if(n<1) return res;
        long R=1;
        for(int i=0;i& res,long R,long number)
    {
        if(R<=number) return ;
        res.push_back(number);
        numbersByRecursion(res,R,number+1);
    }
};
####################################################################
class Solution {
public:
    /**
     * @param n: An integer
     * @return: An array storing 1 to the largest number with n digits.
     */
    vector numbersByRecursion(int n) {
        // write your code here
        vector res;
        if(n<1) return res;
        numbersByRecursion(n,0,res);
        return res;
        
    }
    
private:
    void numbersByRecursion(int n,int ans,vector& res)
    {
        if(n==0)
        {
            if(ans>0)
            {
                res.push_back(ans);
            }
            return ;
        }
        for(int i=0;i<10;i++)
        {
            numbersByRecursion(n-1,ans*10+i,res);
        }
    }
};
##########################################################

public class Solution {
    /**
     * @param n: An integer
     * @return: An array storing 1 to the largest number with n digits.
     */
    public List numbersByRecursion(int n) {
        // write your code here
        ArrayList res=new ArrayList();
        if(n<1)
        {
            return res;
        }
         
        numbersByRecursion(n,0,res);
        return res;
        
    }
    public void numbersByRecursion(int n,int ans,ArrayList res)
    {
        if(n==0)
        {
            if(ans>0)
            {
                res.add(ans);
            }
            return ;
        }
        for(int i=0;i<10;i++)
        {
            numbersByRecursion(n-1,ans*10+i,res);
        }
       
    }
}