#### QUESTION:

You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed).

In one operation, you will create a new array arr, and for each i:

If i % 2 == 0, then arr[i] = perm[i / 2]. If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2]. You will then assign arr to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

**Example 1:**

```
Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1st operation, perm = [0,1]
So it takes only 1 operation.
```

**Example 2:**

```
Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1st operation, perm = [0,2,1,3]
After the 2nd operation, perm = [0,1,2,3]
So it takes only 2 operations.
```

**Example 3:**

```
Input: n = 6
Output: 4
```

**Constraints:**

```
2 <= n <= 1000
n is even.
```

#### EXPLANATION:

看起来是一个中等的题目, 其实只要模拟出来了即可.

#### SOLUTION:

```
class Solution {
func reinitializePermutation(_ n: Int) -> Int {
var perm: [Int] = []
var target: [Int] = []
var result = 0
for i in 0...n-1 {
perm.append(i)
target.append(i)
}
repeat {
var tmpArr = perm
for i in 0...n-1 {
if i % 2 == 0 {
tmpArr[i] = perm[i/2]
} else {
tmpArr[i] = perm[n/2 + (i-1)/2]
}
}
perm = tmpArr
result += 1
} while perm != target
return result
}
}
```