LeetCode 728. Self Dividing Numbers

最新推荐文章于 2020-03-10 21:53:18 发布

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本文介绍了一种查找自除数的高效算法，并提供了一个能在LeetCode上击败100%提交者的C++实现。自除数是指能被其每一位数字整除的数，且不允许包含数字0。文章通过示例详细解释了自除数的概念，并给出了清晰的代码实现。

题目链接：https://leetcode.com/problems/self-dividing-numbers/description/

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.

题目解析：循环一下就好，每个取模得到数位。

代码如下：0ms Accepted beating 100%

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