圆形阵列中相邻元素的最小绝对差

给定n个整数，它们形成一个圆。找到任何相邻对的最小绝对值。如果有许多最佳解决方案，请输出其中任何一个。
注意：它们在圈子中
例子：

Input : arr[] = {10, 12, 13, 15, 10} 
Output : 0
Explanation: |10 - 10| = 0 which is the 
minimum possible.

Input : arr[] = {10, 20, 30, 40}
Output : 10
Explanation: |10 - 20| = 10 which is the 
minimum, 2 3 or 3 4 can be the answers also.  

首先考虑最小值是第一元素和第二元素的最小值。从第二个元素遍历到最后一个。检查每个相邻对的差异并存储最小值。当到达最后一个元素时，请检查其与第一个元素的区别。
下面是上述方法的实现。

C++

// C++ program to find maximum difference
// between adjacent elements in a circular array.
#include 
using namespace std;
 
void minAdjDifference(int arr[], int n)
{
    if (n < 2)
        return;
 
    // Checking normal adjacent elements
    int res = abs(arr[1] - arr[0]);
    for (int i = 2; i < n; i++)
        res = min(res, abs(arr[i] - arr[i - 1]));
 
    // Checking circular link
    res = min(res, abs(arr[n - 1] - arr[0]));
 
    cout << "Min Difference = " << res;
}
 
// driver program to check the above function
int main()
{
    int a[] = { 10, 12, 13, 15, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    minAdjDifference(a, n);
    return 0;
}

Java

// Java program to find maximum difference
// between adjacent elements in a circular
// array.
class GFG {
 
    static void minAdjDifference(int arr[], int n)
    {
        if (n < 2)
            return;
 
        // Checking normal adjacent elements
        int res = Math.abs(arr[1] - arr[0]);
        for (int i = 2; i < n; i++)
            res = Math.min(res, Math.abs(arr[i] - arr[i - 1]));
 
        // Checking circular link
        res = Math.min(res, Math.abs(arr[n - 1] - arr[0]));
 
        System.out.print("Min Difference = " + res);
    }
 
    // driver code
    public static void main(String arg[])
    {
 
        int a[] = { 10, 12, 13, 15, 10 };
        int n = a.length;
 
        minAdjDifference(a, n);
    }
}
 
// This code is contributed by Anant Agarwal
// and improved by Anuj Sharma.

Python3

# Python3 program to find maximum
# difference between adjacent
# elements in a circular array.
 
def minAdjDifference(arr, n):
 
    if (n < 2): return
 
    # Checking normal adjacent elements
    res = abs(arr[1] - arr[0])
     
    for i in range(2, n):
        res = min(res, abs(arr[i] - arr[i - 1]))
 
    # Checking circular link
    res = min(res, abs(arr[n - 1] - arr[0]))
 
    print("Min Difference = ", res)
 
# Driver Code
a = [10, 12, 13, 15, 10]
n = len(a)
minAdjDifference(a, n)
 
# This code is contributed by Anant Agarwal
# and improved by Anuj Sharma.

C#

// C# program to find maximum difference
// between adjacent elements in a circular array.
using System;
 
class GFG {
 
    static void minAdjDifference(int[] arr, int n)
    {
        if (n < 2)
            return;
 
        // Checking normal adjacent elements
        int res = Math.Abs(arr[1] - arr[0]);
        for (int i = 2; i < n; i++)
            res = Math.Min(res, Math.Abs(arr[i] - arr[i - 1]));
 
        // Checking circular link
        res = Math.Min(res, Math.Abs(arr[n - 1] - arr[0]));
 
        Console.Write("Min Difference = " + res);
    }
 
    // driver code
    public static void Main()
    {
        int[] a = { 10, 12, 13, 15, 10 };
        int n = a.Length;
        minAdjDifference(a, n);
    }
}
 
// This code is contributed by Anant Agarwal
// and improved by Anuj Sharma.

Javascript

 
// Javascript program to find maximum difference
// between adjacent elements in a circular array.
 
function minAdjDifference( arr, n)
{
    if (n < 2)
        return;
 
    // Checking normal adjacent elements
    let res = Math.abs(arr[1] - arr[0]);
    for (let i = 2; i < n; i++)
        res = Math.min(res, Math.abs(arr[i] - arr[i - 1]));
 
    // Checking circular link
    res = Math.min(res, Math.abs(arr[n - 1] - arr[0]));
 
    document.write("Min Difference = " + res);
}
 
// driver code
 
    let a = [ 10, 12, 13, 15, 10 ];
    let n = a.length;
    minAdjDifference(a, n);
 

输出：

Min Difference =  0

时间复杂度： O(n)

https://youtu.be/G