LintCode 用递归打印数字

中等 用递归打印数字

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通过

用递归的方法找到从1到最大的N位整数。

您在真实的面试中是否遇到过这个题？ 

Yes

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].

用字符串模拟数字

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> res;        if (n ==0) {            return res;        }        char *num = new char[n+1];        num[n] = '\0';        for (int i = 0; i < 10; ++i) {            num[0] = i + '0';            recursion(num,n,0,res);        }        delete [] num;        return res;    }    void recursion(char * num,int length,int index,vector<int>& res) {        if (index == length-1) {            int a = atoi(num);            if(a != 0) {            res.push_back(a);                }            return;        }        for (int i=0; i< 10; i++) {            num[index + 1] = i +'0';            recursion(num,length, index+1 ,res);                    }        return;    }};