[LeetCode] 728. Self Dividing Numbers

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.

打印出leftright范围内的所有Self Dividing Numbers

1.可以被这个数包含的每个数字整除
2.数字中没有0

拼拼凑凑写出一个粗糙的版本,逻辑混乱,但我第一个版本的代码就长这样,不打草稿多思考就开写就是这个后果

vector<int> selfDividingNumbers(int left, int right) {
    vector<int> result;

    for (int i = left; i <= right; i++)
    {
        if (i < 10)
        {
            result.push_back(i);
            continue;
        }

        if (i % 10 == 0)
        {
            continue;
        }

        bool isSelfDividing = true;

        int tmp = i;

        while (tmp % 10 != 0)
        {
            if (i % (tmp % 10) != 0)
            {
                isSelfDividing = false;
            }
            tmp /= 10;
        }

        if (tmp >= 10)
        {
            isSelfDividing = false;
        }

        if (isSelfDividing)
        {
            result.push_back(i);
        }
    }
    return result;
}

这个版本的代码太乱了,实在拿不出手,优化一下

1.小于10的数都是Self Dividing Numbers
2.不包含0(不能被10整除)
3.能被自身包含的每个数字整除(写个for循环取模)

bool isSelfDividing(int num)
{
    if (num < 10)
    {
        return true;
    }

    for (int i = num; i != 0; i /= 10)
    {
        if (i % 10 == 0)
        {
            return false;
        }
        if (num % (i%10) != 0)
        {
            return false;
        }
    }

    return true;
}

vector<int> selfDividingNumbers(int left, int right) {
    vector<int> result;

    for (int i = left; i <= right; i++)
    {
        if (isSelfDividing(i))
        {
            result.push_back(i);
        }
    }
    return result;
}
posted @ 2018-08-04 16:50  arcsinW  阅读(143)  评论(0编辑  收藏  举报