生成 [1, N] 的排列，其中相邻差的按位异或为 0

给定一个整数N ，任务是生成从 1 到 N 的排列，使得相邻元素之间差异的按位 XOR 为 0，即 | A[1]− A[2] | ^ | A[2]− A[3] | ^ . . . ^ | A[N -1] - A[N] | = 0，其中 |X – Y|表示 X 和 Y 之间的绝对差。

例子：

Input: N = 4
Output: 2 3 1 4
Explanation: |2 -3| ^ |3 -1| ^ |1-4| = 1 ^ 2 ^ 3 = 0

Input: N = 3
Output: 1 2 3

编程需要懂一点英语

方法：这个问题可以根据以下观察来解决：

The XOR of even number of same elements is 0. So if odd number of elements are there (which implies even number of adjacent differences) then arrange them in a way such that the difference between any two adjacent elements is same.
Else if N is even (which implies odd number of adjacent differences) then arrange the first four in such a way that the XOR of first three differences is 0. Then the remaining elements in the above mentioned away.

编程需要懂一点英语

按照下面提到的步骤来实现上述观察：

如果 N 为奇数，则将所有 N 个元素按排序方式排列，因为任意两个相邻元素之间的差为 1，且相邻差的个数为偶数。
如果 N 是偶数：
- 保留 2、3、1、4 作为前四个元素，因为 3 个差异具有 XOR 0。
- 现在从 5 开始并按排序顺序打印剩余的元素，这将使所有剩余的偶数差异的差异为 1。

下面是上述方法的实现。

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to print shuffle array
vector shuffleArray(int n)
{
    vector res;
 
    // Base case
    if (n < 3)
        cout << -1 << endl;
 
    // When n is odd print array in
    // increasing order
    else if (n % 2 != 0) {
        for (int i = 1; i <= n; i++)
            res.push_back(i);
    }
 
    // When n is even print first 2 3 1 4
    // rest element in increasing order
    else {
        res = { 2, 3, 1, 4 };
        for (int i = 5; i <= n; i++)
            res.push_back(i);
    }
    return res;
}
 
// Driver code
int main()
{
    int N = 4;
 
    vector ans = shuffleArray(N);
    for (int x : ans)
        cout << x << " ";
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
public class GFG {
 
  // Function to print shuffle array
  static ArrayList shuffleArray(int n)
  {
    ArrayList res = new ArrayList();
 
    // Base case
    if (n < 3)
      System.out.println(-1);
 
    // When n is odd print array in
    // increasing order
    else if (n % 2 != 0) {
      for (int i = 1; i <= n; i++)
        res.add(i);
    }
 
    // When n is even print first 2 3 1 4
    // rest element in increasing order
    else {
      res.clear();
 
      res.add(2);
      res.add(3);
      res.add(1);
      res.add(4);
 
      for (int i = 5; i <= n; i++)
        res.add(i);
    }
    return res;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int N = 4;
 
    ArrayList ans = shuffleArray(N);
    for (int i = 0; i < ans.size(); i++)
      System.out.print(ans.get(i) + " ");
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 implementation of above approach
 
# Function to print shuffle array
def shuffleArray(n):
 
    res = []
 
    # Base case
    if (n < 3):
        print(-1)
 
    # When n is odd print array in
    # increasing order
    elif (n % 2 != 0):
        for i in range(1, n):
            res.append(i)
 
    # When n is even print first 2 3 1 4
    # rest element in increasing order
    else:
        res = [2, 3, 1, 4]
        for i in range(5, n):
            res.append(i)
 
    return res
 
# Driver code
if __name__ == '__main__':
    n = 4
    res = shuffleArray(n)
    for i in res:
        print(i, end=' ')
 
        # This code is contributed by richasalan57.

C#

// C# implementation of above approach
using System;
using System.Collections;
 
class GFG
{
 
// Function to print shuffle array
static ArrayList shuffleArray(int n)
{
    ArrayList res = new ArrayList();
 
    // Base case
    if (n < 3)
        Console.WriteLine(-1);
 
    // When n is odd print array in
    // increasing order
    else if (n % 2 != 0) {
        for (int i = 1; i <= n; i++)
            res.Add(i);
    }
 
    // When n is even print first 2 3 1 4
    // rest element in increasing order
    else {
        res.Clear();
         
        res.Add(2);
        res.Add(3);
        res.Add(1);
        res.Add(4);
         
        for (int i = 5; i <= n; i++)
            res.Add(i);
    }
    return res;
}
 
// Driver code
public static void Main()
{
    int N = 4;
 
    ArrayList ans = shuffleArray(N);
    foreach (int x in ans)
        Console.Write(x + " ");
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

2 3 1 4

时间复杂度： O(N)
辅助空间： O(1)