给定一个具有n个整数的数组a [],任务是重新排列该数组的元素,以使相邻元素的差值以降序排列。
例子:
Input : arr[] = {1, 2, 3, 4, 5, 6}
Output : 6 1 5 2 4 3
Explanation:
For first two elements the difference is abs(6-1)=5
For next two elements the difference is abs(1-5)=4
For next two elements the difference is abs(5-2)=3
For next two elements the difference is abs(2-4)=2
For next two elements the difference is abs(4-3)=1
Hence, difference array is 5, 4, 3, 2, 1.
Input : arr[] = {7, 10, 2, 4, 5}
Output : 10 2 7 4 5
Explanation:
For first two elements the difference is abs(10-2)=8
For next two elements the difference is abs(2-7)=5
For next two elements the difference is abs(7-4)=3
For next two elements the difference is abs(4-5)=1
Hence, difference array is 8, 5, 3, 1.
方法:
为了解决上述问题,我们必须对数组进行排序,然后按以下顺序为偶数大小的数组打印元素:
a[n], a[1], a[n-1], a[2] ….. a[(n/2)], [(n/2)+1]
对于奇数数组,我们必须按以下顺序打印数组元素:
a[n], a[1], a[n-1], a[2] ….. a[(n/2)+1]
下面是上述方法的实现:
C++
// C++ implementation to Rearrange array
// such that difference of adjacent
// elements is in descending order
#include
using namespace std;
// Function to print array in given order
void printArray(int* a, int n)
{
// Sort the array
sort(a, a + n);
int i = 0;
int j = n - 1;
// Check elements till the middle index
while (i <= j) {
// check if length is odd
// print the middle index at last
if (i == j) {
cout << a[i] << " ";
}
// Print the remaining elements
// in the described order
else {
cout << a[j] << " ";
cout << a[i] << " ";
}
i = i + 1;
j = j - 1;
}
cout << endl;
}
// Driver code
int main()
{
// array declaration
int arr1[] = { 1, 2, 3, 4, 5, 6 };
// size of array
int n1 = sizeof(arr1) / sizeof(arr1[0]);
printArray(arr1, n1);
}
Java
// Java implementation to Rearrange array
// such that difference of adjacent
// elements is in descending order
import java.util.*;
class GFG {
// Function to print array in given order
static void printArray(int []a, int n)
{
// Sort the array
Arrays.sort(a);
int i = 0;
int j = n - 1;
// Check elements till the
// middle index
while (i <= j)
{
// Check if length is odd print
// the middle index at last
if (i == j)
{
System.out.print(a[i] + " ");
}
// Print the remaining elements
// in the described order
else
{
System.out.print(a[j] + " ");
System.out.print(a[i] + " ");
}
i = i + 1;
j = j - 1;
}
System.out.println();
}
// Driver code
public static void main (String[] args)
{
// Array declaration
int arr1[] = { 1, 2, 3, 4, 5, 6 };
// Size of array
int n1 = arr1.length;
printArray(arr1, n1);
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation to rearrange
# array such that difference of adjacent
# elements is in descending order
# Function to print array in given order
def printArray(a, n):
# Sort the array
a.sort();
i = 0;
j = n - 1;
# Check elements till the middle index
while (i <= j):
# Check if length is odd print
# the middle index at last
if (i == j):
print(a[i], end = " ");
# Print the remaining elements
# in the described order
else :
print(a[j], end = " ");
print(a[i], end = " ");
i = i + 1;
j = j - 1;
print();
# Driver code
if __name__ == "__main__" :
# Array declaration
arr1 = [ 1, 2, 3, 4, 5, 6 ];
# Size of array
n1 = len(arr1);
printArray(arr1, n1);
# This code is contributed by AnkitRai01
C#
// C# implementation to rearrange array
// such that difference of adjacent
// elements is in descending order
using System;
class GFG {
// Function to print array in given order
static void printArray(int []a, int n)
{
// Sort the array
Array.Sort(a);
int i = 0;
int j = n - 1;
// Check elements till the
// middle index
while (i <= j)
{
// Check if length is odd print
// the middle index at last
if (i == j)
{
Console.Write(a[i] + " ");
}
// Print the remaining elements
// in the described order
else
{
Console.Write(a[j] + " ");
Console.Write(a[i] + " ");
}
i = i + 1;
j = j - 1;
}
Console.WriteLine();
}
// Driver code
public static void Main (string[] args)
{
// Array declaration
int []arr1 = { 1, 2, 3, 4, 5, 6 };
// Size of array
int n1 = arr1.Length;
printArray(arr1, n1);
}
}
// This code is contributed by AnkitRai01
Javascript
输出:
6 1 5 2 4 3