leetcode -- 526. Beautiful Arrangement 【回溯 + 状态保存 + 状态还原】

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.

解法1(回溯 + 状态保存浪费空间):

public class Solution {

public int countArrangement(int N) {
boolean[] invites = new boolean
;
return find(0,invites,N);
}

public int find(int pos,boolean[] invites,int N){
if(pos == N) return 1;
int sum = 0;
for(int i =0;i < N;i++){
if(!invites[i] && ((i+1) % (pos+1) == 0 || (pos+1) % (i+1) == 0)){
boolean invites2[] = new boolean
;
System.arraycopy(invites, 0, invites2, 0, N);
invites2[i] = true;
sum += find(pos+1,invites2,N);
}
}
return sum;
}}

解法2(回溯 + 状态还原)：
该方法节省了大量的空间，不用每次都新生成数组保存状态。

public class Solution {public static int countArrangement(int N) {
boolean[] invites = new boolean
;
return find(0,invites,N);
}

public static int find(int pos,boolean[] invites,int N){
if(pos == N) return 1;
int sum = 0;
for(int nums =0;nums < N;nums++){
if(!invites[nums] && ((nums+1) % (pos+1) == 0 || (pos+1) % (nums+1) == 0)){
invites[nums] = true;
sum += find(pos+1,invites,N);
invites[nums] = false; //状态还原
}
}
return sum;
}}