leetcode-728. Self Dividing Numbers

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 = 22Output: [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.

题意：

输出 left,right 之间的数，要求是能够被自己的所有位的数整除。

只要一直除自己的每一位的数就可以了，用循环取整和取余就可以得到所有的位数。

自己一开始只想着用String转换，然后用了很多库函数，，果然超时了。。

改为循环就快了很多。。

AC代码：

class Solution {    public List<Integer> selfDividingNumbers(int left, int right) {        List<Integer> list = new ArrayList<>();        for (int i = left;i<=right ;i++ ) {            int j = i;            for (;j > 0 ;j= j/10 ) {                if( (j%10==0 )|| (i%(j%10))!=0) break;            }            if(j==0) list.add(i);        }        return list;    }}

自己一开始写的：(辣眼睛，未AC)

class Solution {    public List<Integer> selfDividingNumbers(int left, int right) {        List<Integer> list = new ArrayList<>();for (int i = left;i <= right ;i++ ) {String temp = String.valueOf(i);int flag = 0;for (int j = 0; j<temp.length() ;i++ ) {if(i % Integer.valueOf(String.valueOf(temp.charAt(j))) != 0){flag = 1;break;}}if(flag == 0){list.add(i);}}return list;    }}