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

该博客探讨了LeetCode中的526题，涉及如何构建一个美丽的排列，使得每个位置上的数要么能整除其位置，要么其位置能整除该数。博客提供了回溯法的解决方案，并特别强调了解决方案如何通过避免状态保存来优化空间效率。

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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 i_th position (1 ≤ i ≤ N) in this array:

The number at the i_th position is divisible by i.
i is divisible by the number at the i_th 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[N];
		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[N];
				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[N];
		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;
	}}