北京工作前一天

今天是2018年1月8日，于沈阳，2018年的第一场大雪，明天就要去北京正式工作了，下午等友人，无聊，故刷Leetcode小题~

第1题、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.

思路：
利用“除10取余”方法，获得整数数位上的每一位数字，然后判断该数字是否为0，是否能被整数整除

class Solution {
    public List<Integer> selfDividingNumbers(int left, int right) {
        List<Integer> list = new LinkedList<Integer>();
        for(int i = left; i <= right; i++) {
            int p = 10, j = i;
            boolean flag = false;
            while(j != 0) {
                int r = j % p;
                if(r == 0 || i % r != 0) {
                    flag = true;
                    break;
                }
                j /= p;
            }       
            if(!flag) {
                list.add(i);
            }
        }
        return list;
    }
}

第2题、693. Binary Number with Alternating Bits

Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:

Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101

Example 2:

Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.

Example 3:

Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.

Example 4:

Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.

思路：
判断一个正整数的二进制是否交替出现0、1，原谅我不要脸地使用了API：Integer.toBinaryString()，直接把整数转化为二进制字符串，然后判断。

class Solution {
    public boolean hasAlternatingBits(int n) {
        String s = Integer.toBinaryString(n);
        int len = s.length();
        boolean flag = true;
        for(int i = 0; i < len - 1; i++) {
            if(s.charAt(i) == s.charAt(i+1)) {
                flag = false;
                break;
            }
        }
        return flag;
    }
}

我就只能写写小题了:-(

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