# 未排序整數列表中最接近的數字 > 原文： [https://www.geeksforgeeks.org/closest-numbers-list-unsorted-integers/](https://www.geeksforgeeks.org/closest-numbers-list-unsorted-integers/) 給定一列不同的未排序整數，找到它們之間的絕對差最小的那對元素？ 如果有多對，請全部找到。 例子： ``` Input : arr[] = {10, 50, 12, 100} Output : (10, 12) The closest elements are 10 and 12 Input : arr[] = {5, 4, 3, 2} Output : (2, 3), (3, 4), (4, 5) ``` 此問題主要是[的擴展。找到任何兩個元素](https://www.geeksforgeeks.org/find-minimum-difference-pair/)之間的最小差異。 1. 排序給定的數組。 2. 在排序數組中找到線性時間中所有對的最小差。 3. 再遍歷排序的數組一次，以步驟 2 中獲得的最小差異打印所有對。 ## C++ ```cpp // CPP program to find minimum difference // an unsorted array. #include<bits/stdc++.h> using namespace std; // Returns minimum difference between any // two pair in arr[0..n-1] void printMinDiffPairs(int arr[], int n) { ???if (n <= 1) ??????return; ???// Sort array elements ???sort(arr, arr+n); ???// Compare differences of adjacent ???// pairs to find the minimum difference. ???int minDiff = arr[1] - arr[0]; ???for (int i = 2 ; i < n ; i++) ??????minDiff = min(minDiff, arr[i] - arr[i-1]); ???// Traverse array again and print all pairs ???// with difference as minDiff. ???for (int i = 1; i < n; i++) ??????if ((arr[i] - arr[i-1]) == minDiff) ?????????cout << "(" << arr[i-1] << ", " ??????????????<< arr[i] << "), "; } // Driver code int main() { ????int arr[] = {5, 3, 2, 4, 1}; ????int n = sizeof(arr) / sizeof(arr[0]); ????printMinDiffPairs(arr, n); ????return 0; } ``` ## Java ```java // Java program to find minimum? // difference an unsorted array. import java.util.*; class GFG? { ??// Returns minimum difference between? ??// any two pair in arr[0..n-1] ??static void printMinDiffPairs(int arr[], int n) ??{ ????if (n <= 1) ??????return; ????// Sort array elements ????Arrays.sort(arr); ????// Compare differences of adjacent ????// pairs to find the minimum difference. ????int minDiff = arr[1] - arr[0]; ????for (int i = 2; i < n; i++) ??????minDiff = Math.min(minDiff, arr[i] - arr[i-1]); ????// Traverse array again and print all pairs ????// with difference as minDiff. ????for ( int i = 1; i < n; i++) ?????{ ????????if ((arr[i] - arr[i-1]) == minDiff) ????????{ ???????????System.out.print("(" + arr[i-1] + ", " ???????????????????????????+ arr[i] + ")," ); ????????}???????????????????? ?????} ??} ??// Driver code ??public static void main (String[] args)? ??{ ????int arr[] = {5, 3, 2, 4, 1}; ????int n = arr.length; ????printMinDiffPairs(arr, n); ??} } // This code is contributed by Ansu Kumari ``` ## Python3 ```py # Python3 program to find minimum? # difference in an unsorted array. # Returns minimum difference between # any two pair in arr[0..n-1] def printMinDiffPairs(arr , n): ????if n <= 1: return ????# Sort array elements ????arr.sort() ????# Compare differences of adjacent ????# pairs to find the minimum difference. ????minDiff = arr[1] - arr[0] ????for i in range(2 , n): ????????minDiff = min(minDiff, arr[i] - arr[i-1]) ????# Traverse array again and print all ????# pairs with difference as minDiff. ????for i in range(1 , n): ????????if (arr[i] - arr[i-1]) == minDiff: ????????????print( "(" + str(arr[i-1]) + ", "? ?????????????????+ str(arr[i]) + "), ", end = '') # Driver code arr = [5, 3, 2, 4, 1] n = len(arr) printMinDiffPairs(arr , n) # This code is contributed by Ansu Kumari ``` ## C# ```cs // C# program to find minimum? // difference an unsorted array. using System; class GFG? { // Returns minimum difference between? // any two pair in arr[0..n-1] static void printMinDiffPairs(int []arr, int n) { ????if (n <= 1) ????return; ????// Sort array elements ????Array.Sort(arr); ????// Compare differences of adjacent ????// pairs to find the minimum difference. ????int minDiff = arr[1] - arr[0]; ????for (int i = 2; i < n; i++) ????minDiff = Math.Min(minDiff, arr[i] - arr[i-1]); ????// Traverse array again and print all pairs ????// with difference as minDiff. ????for ( int i = 1; i < n; i++) ????{ ????????if ((arr[i] - arr[i-1]) == minDiff) ????????{ ????????Console.Write(" (" + arr[i-1] + ", " ??????????????????????????+ arr[i] + "), " ); ????????}????????????? ????} } // Driver code public static void Main ()? { ????int []arr = {5, 3, 2, 4, 1}; ????int n = arr.Length; ????printMinDiffPairs(arr, n); } } // This code is contributed by vt_m ``` ## PHP ```php <?php //PHP program to find minimum difference // an unsorted array. // Returns minimum difference between any // two pair in arr[0..n-1] function printMinDiffPairs($arr, $n) { ????if ($n <= 1) ????????return; ????// Sort array elements ????sort($arr); ????// Compare differences of adjacent ????// pairs to find the minimum? ????// difference. ????$minDiff = $arr[1] - $arr[0]; ????for ($i = 2 ; $i < $n ; $i++) ????????$minDiff = min($minDiff, $arr[$i] ????????????????????????????- $arr[$i-1]); ????// Traverse array again and print all ????// pairs with difference as minDiff. ????for ($i = 1; $i < $n; $i++) ????????if (($arr[$i] - $arr[$i-1]) ==? ????????????????????????????????$minDiff) ????????????echo "(" , $arr[$i-1] , ", ", ????????????????????????$arr[$i] , "), "; } // Driver code ????$arr = array(5, 3, 2, 4, 1); ????$n = sizeof($arr); ????printMinDiffPairs($arr, $n); // This code is contributed by ajit. ?> ``` Output: ``` (1, 2), (2, 3), (3, 4), (4, 5), ``` **上面的程序是否處理重復項？** 上述程序未處理類似{x，x，x}的情況。 在這種情況下，預期輸出（x，x），（x，x），（x，x），但在程序上方會打印（x，x），（x，x） * * * * * *