从给定数组中找到非降序数组

给定一个大小为N / 2的数组A[] ，任务是构造大小为 N 的数组B[]使得：

1. B[] 按非递减顺序排序。
2. A[i] = B[i] + B[n – i + 1]。

注意：数组A[]以这样一种方式给出，即答案总是可能的。
例子：

Input: A[] = {3, 4} 
Output: 0 1 3 3
Input: A[] = {4, 1} 
Output: 0 0 1 4 
 

方法：让我们介绍以下贪心方法。数字将成对恢复(B[0], B[n – 1]) , (B[1], B[n – 2])等等。因此，我们可以对当前对的值进行一些限制（满足有关排序结果的标准）。
最初， l = 0和r = 10 ⁹ ，它们被更新为l = a[i]和r = a[n – i + 1] 。让我在答案中尽可能小。取a[i] = max(l, b[i] – r)和r = b[i] – l ，这样l 的选择方式使得l和r都在限制范围内，而且l也是最小的可能的。
如果l大于我们将向上移动l限制和向下移动r限制，从而为以后的选择留下更少的自由。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Utility function to print
// the contents of the array
void printArr(int b[], int n)
{
    for (int i = 0; i < n; i++)
        cout << b[i] << " ";
}
 
// Function to build array B[]
void ModifiedArray(int a[], int n)
{
    // Lower and upper limits
    int l = 0, r = INT_MAX;
 
    // To store the required array
    int b[n] = { 0 };
 
    // Apply greedy approach
    for (int i = 0; i < n / 2; i++) {
        b[i] = max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }
 
    // Print the built array b[]
    printArr(b, n);
}
 
// Driver code
int main()
{
    int a[] = { 5, 6 };
    int n = sizeof(a) / sizeof(a[0]);
    ModifiedArray(a, 2 * n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class solution
{
    // Utility function to print
    // the contents of the array
    void printArr(int b[], int n)
    {
        for (int i = 0; i < n; i++)
        {
             System.out.print(" " + b[i] + " ");
         }
    }
 
    // Function to build array B[]
    void ModifiedArray(int a[], int n)
    {
        // Lower and upper limits
        int l = 0, r = Integer.MAX_VALUE;
 
        // To store the required array
        int[] b = new int[n];
         
    // Apply greedy approach
    for (int i = 0; i < n / 2; i++) {
        b[i] = Math.max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }
 
    // Print the built array b[]
    printArr(b, n);
}   
// Driver code
public static void main(String args[])
{
   int a[] = { 5, 6 };
   int n = a.length ;
   solution s=new solution();
   s.ModifiedArray(a, 2 * n);
 
}
}
//This code is contributed by Shivi_Aggarwal

Python3

# Python 3 implementation of the approach
import sys
 
# Utility function to print the
# contents of the array
def printArr(b, n):
    for i in range(0, n, 1):
        print(b[i], end = " ")
 
# Function to build array B[]
def ModifiedArray(a, n):
     
    # Lower and upper limits
    l = 0
    r = sys.maxsize
 
    # To store the required array
    b = [0 for i in range(n)]
 
    # Apply greedy approach
    for i in range(0, int(n / 2), 1):
        b[i] = max(l, a[i] - r)
        b[n - i - 1] = a[i] - b[i]
        l = b[i]
        r = b[n - i - 1]
 
    # Print the built array b[]
    printArr(b, n)
 
# Driver code
if __name__ == '__main__':
    a = [5, 6]
    n = len(a)
    ModifiedArray(a, 2 * n)
 
# This code is contributed by
# Shashank_Sharma

C#

// C# implementation of the approach
 
using System;
 
public class GFG{
 
// Utility function to print
// the contents of the array
static void printArr(int []b, int n)
    {
        for (int i = 0; i < n; i++)
        {
            Console.Write(" " + b[i] + " ");
        }
    }
 
    // Function to build array B[]
static    void ModifiedArray(int []a, int n)
    {
        // Lower and upper limits
        int l = 0, r = int.MaxValue;
 
        // To store the required array
        int[] b = new int[n];
         
    // Apply greedy approach
    for (int i = 0; i < n / 2; i++) {
        b[i] = Math.Max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }
 
    // Print the built array b[]
    printArr(b, n);
}
 
        // Driver code
    static public void Main (){
    int []a = { 5, 6 };
    int n = a.Length;
    ModifiedArray(a, 2 * n);
    }
}
// This code is contributed
// by Sach_Code

Javascript

 
// Javascript program of the above approach
 
    // Utility function to prlet
    // the contents of the array
    function prletArr(b, n)
    {
        for (let i = 0; i < n; i++)
        {
             document.write(" " + b[i] + " ");
         }
    }
 
    // Function to build array B[]
    function ModifiedArray(a, n)
    {
        // Lower and upper limits
        let l = 0, r = Number.MAX_VALUE;
 
        // To store the required array
        let b = Array(n).fill(0);
         
    // Apply greedy approach
    for (let i = 0; i < n / 2; i++) {
        b[i] = Math.max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }
 
    // Prlet the built array b[]
    prletArr(b, n);
}
 
// Driver code
 
    let a = [ 5, 6 ];
   let n = a.length ;
   ModifiedArray(a, 2 * n);
     

输出：

0 1 5 5