728. Self Dividing Numbers

最新推荐文章于 2024-07-21 12:49:49 发布

LQCUN

最新推荐文章于 2024-07-21 12:49:49 发布

阅读量118

点赞数

CC 4.0 BY-SA版权

分类专栏：基础

本文链接：https://blog.youkuaiyun.com/u013527178/article/details/79251600

基础专栏收录该内容

50 篇文章

订阅专栏

本文介绍了一种用于找出指定范围内所有自除数的算法。自除数是指能被其每一位数字整除的数，且不含0。文中提供了两种C++实现方法，并详细解释了判断逻辑。

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.

思路：遍历每一个数，判断该数是否是 self-dividing number（核心）。需要注意：因为0不能为分母，所以含0的数一定不是 self-dividing number。

代码1：

class Solution {
public:
    bool isself(int no){
        if(no<10){return true;}
        else {
            string num=to_string(no);
            if(num.find('0')!=std::string::npos){return false;}
            else{
                for(int i=0;i<num.size();i++){
                    int divide=num[i]-'0';
                    if(no%divide!=0){return false;}
                }
            }
        }
        return true;
        }
    vector<int> selfDividingNumbers(int left, int right) {
        vector<int> result;
        for(int i=left;i<=right;i++){
            if(isself(i)){result.push_back(i);}
        }
        return result;
    }
};

代码2：

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