[Swift]LeetCode526. 优美的排列 | Beautiful Arrangement

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10403153.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★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:

1. The number at the ith position is divisible by i.
2. 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:

1. N is a positive integer and will not exceed 15.

---

假设有从 1 到 N 的 N 个整数，如果从这 N 个数字中成功构造出一个数组，使得数组的第 i 位 (1 <= i <= N) 满足如下两个条件中的一个，我们就称这个数组为一个优美的排列。条件：

1. 第 i 位的数字能被 i 整除
2. i 能被第 i 位上的数字整除

现在给定一个整数 N，请问可以构造多少个优美的排列？

示例1:

输入: 2
输出: 2
解释: 

第 1 个优美的排列是 [1, 2]:
  第 1 个位置（i=1）上的数字是1，1能被 i（i=1）整除
  第 2 个位置（i=2）上的数字是2，2能被 i（i=2）整除

第 2 个优美的排列是 [2, 1]:
  第 1 个位置（i=1）上的数字是2，2能被 i（i=1）整除
  第 2 个位置（i=2）上的数字是1，i（i=2）能被 1 整除

说明:

1. N 是一个正整数，并且不会超过15。

---

Runtime: 52 ms

Memory Usage: 18.5 MB

class Solution {
    func countArrangement(_ N: Int) -> Int {
        var nums:[Int] = [Int](repeating:0,count:N)
        for i in 0..<N
        {
            nums[i] = i + 1
        }
        return helper(N, &nums)
    }
    
    func helper(_ n:Int,_ nums:inout [Int]) -> Int
    {
        if n <= 0 {return 1}
        var res:Int = 0
        for i in 0..<n
        {
            if n % nums[i] == 0 || nums[i] % n == 0
            {
                (nums[i], nums[n - 1]) = (nums[n - 1], nums[i])
                res += helper(n - 1, &nums)
                (nums[i], nums[n - 1]) = (nums[n - 1], nums[i])
            }
        }
        return res
    }
}

---

424ms

class Solution {
    func countArrangement(_ N: Int) -> Int {
        var used = [Bool](repeating: false, count: N + 1)
        var result = 0
        backtrack(1, N, &used, &result)
        return result
    }
    
    private func backtrack(_ position: Int, _ n: Int, _ used: inout [Bool], _ result: inout Int) {
        if position > n {
            result += 1
        } else {
            for i in 1...n where !used[i] && (position % i == 0 || i % position == 0) {
                used[i] = true
                backtrack(position + 1, n, &used, &result)
                used[i] = false
            }
        }
    }
}

---

452ms

class Solution {
    var count = 0
    func countArrangement(_ N: Int) -> Int {
        var elements: [Bool] = Array(repeating: false, count: N)
        createNext(elements: &elements, position: 1, n: N)
        return count;
    }
    
    func createNext(elements: inout [Bool], position: Int, n: Int) {
        if position > n {
            count += 1
            return
        }
        for i in 0..<elements.count {
            if !(!elements[i] && ( position % (i+1) == 0 || (i+1) % position == 0)) { continue; }
            elements[i] = true;
            createNext(elements: &elements, position: position+1, n: n)
            elements[i] = false
        }
    }
}

---

536ms

class Solution {
    func countArrangement(_ N: Int) -> Int {
      guard N > 0 else { return 0 }
      
      var visited = Array.init(repeating: 0, count: N + 1)
      var count = 0
      func dfs(position: Int) {
        
        if position > N {
          count += 1
          return
        }
        
        for i in 1..<(N+1) {
          if visited[i] == 0 && (i % position == 0 || position % i == 0) {
            visited[i] = 1
            dfs(position: position + 1)
            visited[i] = 0
          }
        }
        
      }
      
      dfs(position: 1)
      return count
    }
}

---

820ms

class Solution {
    func countArrangement(_ N: Int) -> Int {
        var num = 0
        var mark = Array(repeatElement(0, count: N + 1))
        insertNum(N: N, index: 1, mark: &mark, result: &num)
        return num
    }
    
    func insertNum(N: Int, index: Int, mark: inout [Int], result: inout Int) {
        if N == index - 1 {
            result += 1
            return
        }
        
        for j in 1...N {
            if mark[j] == 0 && (j % index == 0 || index % j == 0) {
                mark[j] = 1
                insertNum(N: N, index: index + 1, mark: &mark, result: &result)
                mark[j] = 0
            }
        }
    }
}