leetcode 526. Beautiful Arrangement 递归实现全排列 + 经典深度优先遍历DFS做法

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.

本题题意很简单，直接深度优先遍历DFS即可

建议和leetcode 667. Beautiful Arrangement II 双指针遍历

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>

using namespace std;

class Solution
{
public:
int count = 0;
int countArrangement(int n)
{
vector<bool> visit(n + 1, false);
dfs(visit, 1, n);
return count;
}

void dfs(vector<bool>& visit, int index, int size)
{
if (index >= size+1)
{
count++;
return;
}
else
{
for (int i = 1; i <= size; i++)
{
if (visit[i] == false && (index % i == 0 || i % index == 0))
{
visit[i] = true;
dfs(visit, index + 1, size);
visit[i] = false;
}
}
}
}
};