Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:
The number at the ith position is divisible by i.
i is divisible by the number at the ith position.
Now given N, how many beautiful arrangements can you construct?
Example 1:
Input: 2
Output: 2
Explanation:
The first beautiful arrangement is [1, 2]:
Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
The second beautiful arrangement is [2, 1]:
Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.
Note:
N is a positive integer and will not exceed 15.
思路：
从第一位到第N位，我们都要找到对应的没有放置过的数字来放，每一位都会有很多个数字可以放，而放了之后以后就不能放了，这样一直放到最后一位，如果都能放到数字，那就是一种漂亮的安排，总结果就要加一。
可以通过递归来实现，每一次递归我们都判断当前位置有哪些没放过的数字可以放，对于数字有没有放过我们需要一个数字来记录。对于每个放在这一位的数字，都是一种可能性，我们要继续往后递归看能不能全部放完才能知道要不要算作一种。如果所有都放完了那就算作一种了，总结过可以加一。
要注意这里的位置并不是从0开始算的，而是1。
class Solution
public int countArrangement(int N) {
int[] num = new int[N];
int res = findWay(num, 1);
return res;
}
public int findWay(int[] num, int index) {
if (index == num.length + 1)
return 1;
int total = 0;
for (int i = 0; i < num.length; i++) {
if (num[i] != 1) {
if ((i + 1) % index == 0 || index % (i + 1) == 0) {
int[] newNum = num.clone();
newNum[i] = 1;
total += findWay(newNum, index + 1);
}
}
}
return
