通过用和和差交替替换第一个和中间元素来减少数组

给定一个大小为N的数组arr[] ，任务是在连续删除数组的^第一个和中间元素并在数组末尾附加它们的和和差后找到数组的最后一个剩余元素。

例子：

Input: A = {2, 4, 1, 5, 7}
Output: 5
Explanation: During the 1st iteration, remove arr[0] and arr[2] from the array 
and append their sum (2 + 1 = 3) to the end of the array. 
Hence, arr[] = {4, 5, 7, 3}. 
Again repeat the same process and remove arr[0] and arr[2] from the array 
and now append their difference (7 – 4) = 3. 
Hence, arr[] = {5, 3, 3}.
 After the next iteration, array will become arr[] = {3, 8}. 
And finally after the last iteration, arr[] = {5} which is the last required value.

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 15

方法：给定的问题是一个基于实现的问题，可以通过重复执行给定的步骤来解决：

• 维护一个变量op ，它存储操作的计数。
• 删除数组的第一个和中间元素，直到它的大小大于1 ，并根据op将它们的加法或减法附加到数组中（即，如果 op 是偶数，则执行加法，否则执行减法）。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to find the last element
// in array after the given operations
int lastElement(vector& arr)
{
 
    // Maintain operation count
    int op = 0, x = 0;
 
    // Loop until array has more
    // than 1 element in it
    while (arr.size() != 1) {
 
        // Stores middle index
        int mid = arr.size() / 2;
 
        // For even iterations perform
        // addition otherwise subtraction
        if (op % 2 == 0) {
            x = arr[mid] + arr[0];
            arr.erase(arr.begin() + mid);
            arr.erase(arr.begin());
        }
        else {
            x = arr[mid] - arr[0];
            arr.erase(arr.begin() + mid);
            arr.erase(arr.begin());
        }
 
        // Append in the end
        arr.push_back(x);
 
        // Increment operation count
        op += 1;
    }
 
    // Return Answer
    return arr[0];
}
 
// Driver Code
int main()
{
    vector arr = { 2, 4, 1, 5, 7 };
    cout << lastElement(arr);
 
    return 0;
}
 
    // This code is contributed by rakeshsahni

Java

// Java program of the above approach
import java.util.*;
 
class GFG{
 
// Function to find the last element
// in array after the given operations
static int lastElement(Vector arr)
{
 
    // Maintain operation count
    int op = 0, x = 0;
 
    // Loop until array has more
    // than 1 element in it
    while (arr.size() != 1) {
 
        // Stores middle index
        int mid = arr.size() / 2;
 
        // For even iterations perform
        // addition otherwise subtraction
        if (op % 2 == 0) {
            x = arr.get(mid) + arr.get(0);
            arr.remove(mid);
            arr.remove(0);
        }
        else {
            x = arr.get(mid) - arr.get(0);
            arr.remove(mid);
            arr.remove(0);
        }
 
        // Append in the end
        arr.add(x);
 
        // Increment operation count
        op += 1;
    }
 
    // Return Answer
    return arr.get(0);
}
 
// Driver Code
public static void main(String[] args)
{
    Integer []arr = { 2, 4, 1, 5, 7 };
    Vector v = new Vector();
    Collections.addAll(v, arr);
    System.out.print(lastElement(v));
 
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program of the above approach
 
# Function to find the last element
# in array after the given operations
def lastElement(arr):
 
    # Maintain operation count
    op = 0
 
    # Loop until array has more
    # than 1 element in it
    while len(arr) != 1:
 
        # Stores middle index
        mid = len(arr)//2
 
        # For even iterations perform
        # addition otherwise subtraction
        if op % 2 == 0:
            x = arr.pop(mid) + arr.pop(0)
        else:
            x = arr.pop(mid) - arr.pop(0)
 
        # Append in the end
        arr.append(x)
 
        # Increment operation count
        op += 1
 
    # Return Answer
    return arr[0]
 
 
# Driver Code
if __name__ == "__main__":
    arr = [2, 4, 1, 5, 7]
    print(lastElement(arr))

C#

// C# program of the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Function to find the last element
  // in array after the given operations
  static int lastElement(List arr)
  {
 
    // Maintain operation count
    int op = 0, x = 0;
 
    // Loop until array has more
    // than 1 element in it
    while (arr.Count != 1) {
 
      // Stores middle index
      int mid = arr.Count / 2;
 
      // For even iterations perform
      // addition otherwise subtraction
      if (op % 2 == 0) {
        x = arr[mid] + arr[0];
        arr.RemoveAt(mid);
        arr.RemoveAt(0);
      }
      else {
        x = arr[mid] - arr[0];
        arr.RemoveAt(mid);
        arr.RemoveAt(0);
      }
 
      // Append in the end
      arr.Add(x);
 
      // Increment operation count
      op += 1;
    }
 
    // Return Answer
    return arr[0];
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 2, 4, 1, 5, 7 };
    List v = new List(arr);
    Console.Write(lastElement(v));
 
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

       // JavaScript code for the above approach
 
       // Function to find the last element
       // in array after the given operations
       function lastElement(arr) {
 
           // Maintain operation count
           let op = 0, x = 0;
 
           // Loop until array has more
           // than 1 element in it
           while (arr.length != 1) {
 
               // Stores middle index
               let mid = Math.floor(arr.length / 2);
 
               // For even iterations perform
               // addition otherwise subtraction
               if (op % 2 == 0) {
                   x = arr[mid] + arr[0];
                   let p1 = arr.slice(1, mid)
                   let p2 = arr.slice(mid + 1)
                   arr = [...p1.concat(p2)];
               }
               else {
                   x = arr[mid] - arr[0];
                   let p1 = arr.slice(1, mid)
                   let p2 = arr.slice(mid + 1)
                   arr = [...p1.concat(p2)];
 
 
               }
 
               // Append in the end
               arr.push(x);
 
               // Increment operation count
               op = op + 1;
           }
 
           // Return Answer
           return arr[0];
       }
 
       // Driver Code
 
       let arr = [2, 4, 1, 5, 7];
       document.write(lastElement(arr));
 
    // This code is contributed by Potta Lokesh
   

输出

5

时间复杂度： O(N ² )
辅助空间： O(1)