未排序整数列表中最接近的数字

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?如果有多个对，则将它们全部找到。
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“查找任何两个元素之间的最小差异”的扩展。

1. 对给定的数组进行排序。
2. 在排序的数组中找到线性时间中所有对的最小差。
3. 再遍历排序后的数组一次，以步骤2中获得的最小差异打印所有对。

C++

// CPP program to find minimum difference
// an unsorted array.
#include
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 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

# 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#

// 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

Javascript

 
// JavaScript 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
    arr.sort();
       
    // Compare differences of adjacent
    // pairs to find the minimum difference.
    let minDiff = arr[1] - arr[0];
    for (let 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 ( let i = 1; i < n; i++)
     {
        if ((arr[i] - arr[i-1]) == minDiff)
        {
           document.write("(" + arr[i-1] + ", "
                           + arr[i] + ") , " );
        }                   
     }
  }
       
 
// Driver code
         
        let arr = [5, 3, 2, 4, 1];
    let n = arr.length;
    printMinDiffPairs(arr , n);
 

输出：

(1, 2), (2, 3), (3, 4), (4, 5), 

上面的程序是否处理重复项?
上述程序无法处理{x，x，x}之类的情况。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)