给定一个由N个整数组成的数组，任务是将它们以圆形排列方式进行布置，以使该元素严格小于其相邻元素的总和。如果这样的安排是不可能的，则打印-1 。
注意，可以有多种方式来布置元素，从而满足条件，并且任务是找到任何这样的布置。
例子：
Input: arr[] = {1, 4, 4, 3, 2}
Output: 1 3 4 4 2
arr[0] = 1 < (2 + 3)
arr[1] = 4 < (1 + 4)
arr[2] = 4 < (4 + 3)
arr[3] = 3 < (4 + 2)
arr[4] = 2 < (3 + 1)
Input: arr[] = {8, 13, 5}
Output: -1
方法：可以使用贪婪方法解决问题，我们首先对数组进行排序，然后将最小的元素放在开头，第二个最小放在结尾，第三个最小放在第二个位置，第四个放在倒数第二个位置在另一个数组中。安排完成后，检查是否满足给定条件。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the arrangement that
// satisifes the given condition
void printArrangement(int a[], int n)
{
// Sort the array initially
sort(a, a + n);
// Array that stores the arrangement
int b[n];
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
int low = 0, high = n - 1;
for (int i = 0; i < n; i++) {
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (int i = 0; i < n; i++) {
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0) {
if (b[n - 1] + b[1] <= b[i]) {
cout << -1;
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1)) {
if (b[n - 2] + b[0] <= b[i]) {
cout << -1;
return;
}
}
else {
if (b[i - 1] + b[i + 1] <= b[i]) {
cout << -1;
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (int i = 0; i < n; i++)
cout << b[i] << " ";
}
// Driver code
int main()
{
int a[] = { 1, 4, 4, 3, 2 };
int n = sizeof(a) / sizeof(a[0]);
printArrangement(a, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function to print the arrangement that
// satisifes the given condition
static void printArrangement(int a[], int n)
{
// Sort the array initially
Arrays.sort(a);
// Array that stores the arrangement
int b[] = new int[n];
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
int low = 0, high = n - 1;
for (int i = 0; i < n; i++)
{
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (int i = 0; i < n; i++)
{
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0)
{
if (b[n - 1] + b[1] <= b[i])
{
System.out.print(-1);
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1))
{
if (b[n - 2] + b[0] <= b[i])
{
System.out.print(-1);
return;
}
}
else
{
if (b[i - 1] + b[i + 1] <= b[i])
{
System.out.print(-1);
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (int i = 0; i < n; i++)
System.out.print(b[i] + " ");
}
// Driver code
public static void main (String[] args)
{
int a[] = { 1, 4, 4, 3, 2 };
int n = a.length;
printArrangement(a, n);
}
}
// This code is contributed by anuj_67..Python3
# Python3 implementation of the approach
# Function to print the arrangement that
# satisifes the given condition
def printArrangement(a, n):
# Sort the array initially
a = sorted(a)
# Array that stores the arrangement
b = [0 for i in range(n)]
# Once the array is sorted
# Re-fill the array again in the
# mentioned way in the approach
low = 0
high = n - 1
for i in range(n):
if (i % 2 == 0):
b[low] = a[i]
low += 1
else:
b[high] = a[i]
high -= 1
# Iterate in the array
# and check if the arrangement made
# satisfies the given condition or not
for i in range(n):
# For the first element
# the adjacents will be a[1] and a[n-1]
if (i == 0):
if (b[n - 1] + b[1] <= b[i]):
print("-1")
return
# For the last element
# the adjacents will be a[0] and a[n-2]
elif (i == (n - 1)) :
if (b[n - 2] + b[0] <= b[i]):
print("-1")
return
else:
if (b[i - 1] + b[i + 1] <= b[i]):
print("-1")
return
# If we reach this position then
# the arrangement is possible
for i in range(n):
print(b[i], end = " ")
# Driver code
a = [ 1, 4, 4, 3, 2 ]
n = len(a)
printArrangement(a, n)
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the arrangement that
// satisifes the given condition
static void printArrangement(int []a, int n)
{
// Sort the array initially
Array.Sort(a);
// Array that stores the arrangement
int []b = new int[n];
// Once the array is sorted
// Re-fill the array again in the
// mentioned way in the approach
int low = 0, high = n - 1;
for (int i = 0; i < n; i++)
{
if (i % 2 == 0)
b[low++] = a[i];
else
b[high--] = a[i];
}
// Iterate in the array
// and check if the arrangement made
// satisfies the given condition or not
for (int i = 0; i < n; i++)
{
// For the first element
// the adjacents will be a[1] and a[n-1]
if (i == 0)
{
if (b[n - 1] + b[1] <= b[i])
{
Console.Write(-1);
return;
}
}
// For the last element
// the adjacents will be a[0] and a[n-2]
else if (i == (n - 1))
{
if (b[n - 2] + b[0] <= b[i])
{
Console.Write(-1);
return;
}
}
else
{
if (b[i - 1] + b[i + 1] <= b[i])
{
Console.Write(-1);
return;
}
}
}
// If we reach this position then
// the arrangement is possible
for (int i = 0; i < n; i++)
Console.Write(b[i] + " ");
}
// Driver code
public static void Main ()
{
int []a = { 1, 4, 4, 3, 2 };
int n = a.Length;
printArrangement(a, n);
}
}
// This code is contributed by anuj_67..Javascript
输出：
1 3 4 4 2时间复杂度： O(N log N)
辅助空间： O(n)