生成一个给定大小的数组，该数组具有相等的计数以及奇数和偶数之和

给定一个整数N ，任务是找到一个长度为N的数组，该数组包含相同数量的奇数和偶数元素，且该数组中的偶数和奇数元素之和相等。
注意：如果没有这样的数组，则打印-1。
例子：

Input: N = 4 
Output: 1 2 5 4 
Explanation: 
Even elements of the array – {2, 4}, S(even) = 6 
Odd elements of the array – {1, 5}, S(odd) = 6
Input: N = 6 
Output: -1 
Explanation: 
There are no such array which contains 3 even elements and 3 odd elements with equal sum. 
 

方法：该问题的主要观察结果是，只有数组的长度是4的倍数才能形成具有相等数量的偶数和奇数元素且具有相等总和的数组。下面是步骤说明：

• 数组的偶数元素是从2开始的自然数的前N / 2个偶数元素。
• 类似地，(N / 2 – 1)所述阵列的奇数元素是第一(N / 2 – 1)从1开始自然数的奇数元素。
• 数组的最后一个奇数元素是使数组的偶数和奇数元素之和相等的必需值。

  Last Odd Element = 
     (sum of even elements) - 
     (sum of N/2 - 1 odd elements)

  下面是上述方法的实现：

  C++

  // C++ implementation to find the
  // array containing same count of
  // even and odd elements with equal
  // sum of even and odd elements
    
  #include 
    
  using namespace std;
    
  // Function to find the array such that
  // the array contains the same count
  // of even and odd elements with equal
  // sum of even and odd elements
  void findSolution(int N)
  {
    
      // Length of array which is not
      // divisible by 4 is unable to
      // form such array
      if (N % 4 != 0)
          cout << -1 << "\n";
      else {
          int temp = 0, sum_odd = 0,
              sum_even = 0;
          int result[N] = { 0 };
    
          // Loop to find the resulted
          // array containing the same
          // count of even and odd elements
          for (int i = 0; i < N; i += 2) {
              temp += 2;
    
              result[i + 1] = temp;
              // Find the total sum
              // of even elements
              sum_even += result[i + 1];
    
              result[i] = temp - 1;
              // Find the total sum
              // of odd elements
              sum_odd += result[i];
          }
    
          // Find the difference between the
          // total sum of even and odd elements
          int diff = sum_even - sum_odd;
    
          // The difference will be added
          // in the last odd element
          result[N - 2] += diff;
    
          for (int i = 0; i < N; i++)
              cout << result[i] << " ";
          cout << "\n";
      }
  }
    
  // Driver Code
  int main()
  {
      int N = 8;
      findSolution(N);
      return 0;
  }

  Java

  // Java implementation to find the
  // array containing same count of
  // even and odd elements with equal
  // sum of even and odd elements
    
  class GFG{
    
  // Function to find the array such that
  // the array contains the same count
  // of even and odd elements with equal
  // sum of even and odd elements
  static void findSolution(int N)
  {
    
      // Length of array which is not
      // divisible by 4 is unable to
      // form such array
      if (N % 4 != 0)
          System.out.print(-1 + "\n");
    
      else 
      {
          int temp = 0, sum_odd = 0;
          int sum_even = 0;
          int result[] = new int[N];
    
          // Loop to find the resulted
          // array containing the same
          // count of even and odd elements
          for(int i = 0; i < N; i += 2)
          {
             temp += 2;
             result[i + 1] = temp;
               
             // Find the total sum
             // of even elements
             sum_even += result[i + 1];
             result[i] = temp - 1;
               
             // Find the total sum
             // of odd elements
             sum_odd += result[i];
          }
            
          // Find the difference between the
          // total sum of even and odd elements
          int diff = sum_even - sum_odd;
    
          // The difference will be added
          // in the last odd element
          result[N - 2] += diff;
    
          for(int i = 0; i < N; i++)
             System.out.print(result[i] + " ");
          System.out.print("\n");
      }
  }
    
  // Driver Code
  public static void main(String[] args)
  {
      int N = 8;
      findSolution(N);
  }
  }
    
  // This code is contributed by Amit Katiyar

  Python3

  # Python3 implementation to find the 
  # array containing same count of 
  # even and odd elements with equal 
  # sum of even and odd elements 
    
  # Function to find the array such that 
  # the array contains the same count 
  # of even and odd elements with equal 
  # sum of even and odd elements 
  def findSolution(N): 
    
      # Length of array which is not 
      # divisible by 4 is unable to 
      # form such array 
      if (N % 4 != 0): 
          print(-1) 
      else: 
          temp = 0
          sum_odd = 0
          sum_even = 0
          result = [0] * N 
    
          # Loop to find the resulted 
          # array containing the same 
          # count of even and odd elements 
          for i in range(0, N, 2): 
              temp += 2
              result[i + 1] = temp 
                
              # Find the total sum 
              # of even elements 
              sum_even += result[i + 1] 
              result[i] = temp - 1
                
              # Find the total sum 
              # of odd elements 
              sum_odd += result[i] 
    
          # Find the difference between the 
          # total sum of even and odd elements 
          diff = sum_even - sum_odd 
    
          # The difference will be added 
          # in the last odd element 
          result[N - 2] += diff 
            
          for i in range(N): 
              print(result[i], end = " ") 
          print() 
    
  # Driver Code 
  N = 8; 
  findSolution(N)
        
  # This code is contributed by divyamohan123

  C#

  // C# implementation to find the
  // array containing same count of
  // even and odd elements with equal
  // sum of even and odd elements
  using System;
    
  class GFG{
    
  // Function to find the array such that
  // the array contains the same count
  // of even and odd elements with equal
  // sum of even and odd elements
  static void findSolution(int N)
  {
    
      // Length of array which is not
      // divisible by 4 is unable to
      // form such array
      if (N % 4 != 0)
          Console.Write(-1 + "\n");
    
      else
      {
          int temp = 0, sum_odd = 0;
          int sum_even = 0;
          int []result = new int[N];
    
          // Loop to find the resulted
          // array containing the same
          // count of even and odd elements
          for(int i = 0; i < N; i += 2)
          {
             temp += 2;
             result[i + 1] = temp;
               
             // Find the total sum
             // of even elements
             sum_even += result[i + 1];
             result[i] = temp - 1;
               
             // Find the total sum
             // of odd elements
             sum_odd += result[i];
          }
            
          // Find the difference between the
          // total sum of even and odd elements
          int diff = sum_even - sum_odd;
    
          // The difference will be added
          // in the last odd element
          result[N - 2] += diff;
    
          for(int i = 0; i < N; i++)
             Console.Write(result[i] + " ");
          Console.Write("\n");
      }
  }
    
  // Driver Code
  public static void Main(String[] args)
  {
      int N = 8;
      findSolution(N);
  }
  }
    
  // This code is contributed by Rohit_ranjan

  Javascript

    
  // JavaScript implementation to find the
  // array containing same count of
  // even and odd elements with equal
  // sum of even and odd elements
    
    
  // Function to find the array such that
  // the array contains the same count
  // of even and odd elements with equal
  // sum of even and odd elements
  function findSolution(N)
  {
    
      // Length of array which is not
      // divisible by 4 is unable to
      // form such array
      if (N % 4 != 0)
          document.write(-1 + "<br>");
      else {
          let temp = 0, sum_odd = 0,
              sum_even = 0;
          let result = new Uint8Array(N);
    
          // Loop to find the resulted
          // array containing the same
          // count of even and odd elements
          for (let i = 0; i < N; i += 2) {
              temp += 2;
    
              result[i + 1] = temp;
              // Find the total sum
              // of even elements
              sum_even += result[i + 1];
    
              result[i] = temp - 1;
              // Find the total sum
              // of odd elements
              sum_odd += result[i];
          }
    
          // Find the difference between the
          // total sum of even and odd elements
          let diff = sum_even - sum_odd;
    
          // The difference will be added
          // in the last odd element
          result[N - 2] += diff;
    
          for (let i = 0; i < N; i++)
              document.write(result[i] + " ");
          document.write("<br>");
      }
  }
    
  // Driver Code
      let N = 8;
      findSolution(N);
    
    
    
  // This code is contributed by Surbhi Tyagi.

  输出：

  1 2 3 4 5 6 11 8

  性能分析：

  • 时间复杂度： O(N)
  • 辅助空间： O(1)

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