[LeetCode] 728. Self Dividing Numbers

[grandyang]

 

A  self-dividing number  is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

Example 1:

Input: 
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

 

Note:

• The boundaries of each input argument are 1 <= left <= right <= 10000.

 

这道题让我们找一个给定范围内的所有的自整除数字，所谓的自整除数字就是该数字可以整除其每一个位上的数字。既然这道题是Easy类，那么一般来说不需要用tricky的方法，直接暴力搜索就行了，遍历区间内的所有数字，然后调用子函数判断其是否是自整除数，是的话就加入结果res中。在子函数中，我们先把数字转为字符串，然后遍历每个字符，只要其为0，或者num无法整除该位上的数字，就返回false，循环结束后返回true，参见代码如下：

 

解法一：

class Solution {
public:
    vector<int> selfDividingNumbers(int left, int right) {
        vector<int> res;
        for (int i = left; i <= right; ++i) {
            if (check(i)) res.push_back(i);
        }
        return res;
    }
    bool check(int num) {
        string str = to_string(num);
        for (char c : str) {
            if (c == '0' || num % (c - '0')) return false;
        }
        return true;
    }
};

 

我们可以不用子函数，直接在大的for循环中加上一个for循环进行判断即可，参见代码如下：

 

解法二：

class Solution {
public:
    vector<int> selfDividingNumbers(int left, int right) {
        vector<int> res;
        for (int i = left, n = 0; i <= right; ++i) {
            for (n = i; n > 0; n /= 10) {
                if (n % 10 == 0 || i % (n % 10) != 0) break;
            }
            if (n == 0) res.push_back(i);
        }
        return res;
    }
};

 

类似题目：

Perfect Number

 

参考资料：

https://discuss.leetcode.com/topic/111201/java-c-clean-code

 

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