圆形数组中相邻元素的最小绝对差 · GeeksForGeeks 中文系列教程

# 圆形数组中相邻元素的最小绝对差 > 原文： [https://www.geeksforgeeks.org/minimum-absolute-difference-adjacent-elements-circular-array/](https://www.geeksforgeeks.org/minimum-absolute-difference-adjacent-elements-circular-array/) 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和第二元素。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，检查其与第一个元素的区别。下面是上述方法的实现。 ## C++ ```cpp // C++ program to find maximum difference // between adjacent elements in a circular array. #include <bits/stdc++.h> 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 // 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 ```py # 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# ```cs // 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. ``` ## PHP ```php <?php // PHP program to find maximum // difference between adjacent // elements in a circular array. function minAdjDifference($arr, $n) { if ($n < 2) return; // Checking normal // adjacent elements $res = abs($arr[1] - $arr[0]); for ($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])); echo "Min Difference = ", $res; } // Driver Code $a = array(10, 12, 13, 15, 10); $n = count($a); minAdjDifference($a, $n); //This code is contributed by anuj_67 //and improved by Anuj Sharma. ?> ``` Output: ``` Min Difference = 0 ``` **时间复杂度**：`O(n)` 本文由 [**努力者**](https://www.facebook.com/raja.vikramaditya.7) 贡献。如果您喜欢 GeeksforGeeks 并希望做出贡献，则还可以使用 [tribution.geeksforgeeks.org](http://www.contribute.geeksforgeeks.org) 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。查看您的文章出现在 GeeksforGeeks 主页上，并帮助其他 Geeks。