数组中两个元素之间的最大差值
给定一个由 N 个整数组成的数组 arr[] ，任务是找出数组中任意两个元素之间的最大差值。 例:
输入: arr[] = {2，1，5，3} 输出:4 | 5–1 | = 4 输入: arr[] = {-10，4，-9，-5} 输出: 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。 以下是上述办法的实施情况:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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));
}
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the maximum
// absolute difference between
// any two elements of the array
function maxAbsDiff($arr, $n)
{
// To store the minimum and the maximum
// elements from the array
$minEle = $arr[0];
$maxEle = $arr[0];
for ($i = 1; $i < $n; $i++)
{
$minEle = min($minEle, $arr[$i]);
$maxEle = max($maxEle, $arr[$i]);
}
return ($maxEle - $minEle);
}
// Driver code
$arr = array(2, 1, 5, 3);
$n = sizeof($arr);
echo maxAbsDiff($arr, $n);
// This code is contributed
// by Akanksha Rai
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the maximum
// absolute difference between
// any two elements of the array
function maxAbsDiff(arr, n)
{
// To store the minimum and the maximum
// elements from the array
let minEle = arr[0];
let maxEle = arr[0];
for (let i = 1; i < n; i++) {
minEle = Math.min(minEle, arr[i]);
maxEle = Math.max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
let arr = [ 2, 1, 5, 3 ];
let n = arr.length;
document.write(maxAbsDiff(arr, n));
</script>
Output
4
时间复杂度 : O(n)
辅助空间 : O(1)
另一种方法(使用 STL) :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:
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum
// absolute difference between
// any two elements of the array
int maxAbsDiff(int arr[], int n)
{
// To find the minimum and the maximum element
// using stl
int maxele = *max_element(arr, arr + n);
int minele = *min_element(arr, arr + n);
// make variable to store answer
int ans = abs(maxele - minele);
return ans;
}
// Driver code
int main()
{
int arr[] = { -10, 4, -9, -5 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
}
Output
14
时间复杂度 : O(n)
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