未排序整数列表中最接近的数字 · GeeksForGeeks 中文系列教程

# 未排序整数列表中最接近的数字 > 原文： [https://www.geeksforgeeks.org/closest-numbers-list-unsorted-integers/](https://www.geeksforgeeks.org/closest-numbers-list-unsorted-integers/) 给定一列不同的未排序整数，找到它们之间的绝对差最小的那对元素？如果有多对，请全部找到。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) ``` 此问题主要是[的扩展。找到任何两个元素](https://www.geeksforgeeks.org/find-minimum-difference-pair/)之间的最小差异。 1. 排序给定的数组。 2. 在排序数组中找到线性时间中所有对的最小差。 3. 再遍历排序的数组一次，以步骤 2 中获得的最小差异打印所有对。 ## 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), ``` **上面的程序是否处理重复项？** 上述程序未处理类似{x，x，x}的情况。在这种情况下，预期输出（x，x），（x，x），（x，x），但在程序上方会打印（x，x），（x，x） * * * * * *