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：问题中的关键观察结果是，只有数组长度为4的倍数才能形成具有相等数目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 KatiyarPython3
# 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 divyamohan123C#
// 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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