给定N个整数的数组arr [] ,任务是查找数组中任何两个元素之间的最大差。
例子:
Input: arr[] = {2, 1, 5, 3}
Output: 4
|5 – 1| = 4
Input: arr[] = {-10, 4, -9, -5}
Output: 14
方法:数组中的最大绝对差将始终是数组中最小和最大元素之间的绝对差。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum
// absolute difference between
// any two elements of the array
int maxAbsDiff(int arr[], int n)
{
// To store the minimum and the maximum
// elements from the array
int minEle = arr[0];
int maxEle = arr[0];
for (int i = 1; i < n; i++) {
minEle = min(minEle, arr[i]);
maxEle = max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
int main()
{
int arr[] = { 2, 1, 5, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to return the maximum
// absolute difference between
// any two elements of the array
static int maxAbsDiff(int arr[], int n)
{
// To store the minimum and the maximum
// elements from the array
int minEle = arr[0];
int maxEle = arr[0];
for (int i = 1; i < n; i++) {
minEle = Math.min(minEle, arr[i]);
maxEle = Math.max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 2, 1, 5, 3 };
int n = arr.length;
System.out.print(maxAbsDiff(arr, n));
}
}
Python3
# Python3 implementation of the approach
# Function to return the maximum
# absolute difference between
# any two elements of the array
def maxAbsDiff(arr, n):
# To store the minimum and the maximum
# elements from the array
minEle = arr[0]
maxEle = arr[0]
for i in range(1, n):
minEle = min(minEle, arr[i])
maxEle = max(maxEle, arr[i])
return (maxEle - minEle)
# Driver code
arr = [2, 1, 5, 3]
n = len(arr)
print(maxAbsDiff(arr, n))
# This code is contributed
# by mohit kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum
// absolute difference between
// any two elements of the array
static int maxAbsDiff(int []arr, int n)
{
// To store the minimum and the maximum
// elements from the array
int minEle = arr[0];
int maxEle = arr[0];
for (int i = 1; i < n; i++)
{
minEle = Math.Min(minEle, arr[i]);
maxEle = Math.Max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
public static void Main()
{
int[] arr = { 2, 1, 5, 3 };
int n = arr.Length;
Console.WriteLine(maxAbsDiff(arr, n));
}
}
// This code is contributed by Ryuga
PHP
输出:
4