数组中的正负值对

给定一个由不同整数组成的数组，打印所有具有数组中存在的数字的正值和负值的对。我们需要按出现的顺序打印对。应该首先打印任何元素首先出现的对。

例子：

Input  :  arr[] = { 1, -3, 2, 3, 6, -1 }
Output : -1 1 -3 3

Input  :  arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }
Output : -1 1 -4 4 -8 8 -9 9

方法一（简单：O(n ² )）
这个想法是使用两个嵌套循环。对于每个元素 arr[i]，从索引 i + 1 到 n – 1 中找到 arr[i] 的负数并将其存储在另一个数组中。对于输出，对存储的元素进行排序并打印存储元素的负正值。

下面是这个方法的实现：

C++

// Simple CPP program to find pairs of positive
// and negative values present in an array.
#include 
using namespace std;
 
// Print pair with negative and positive value
void printPairs(int arr[], int n)
{
    vector v;
 
    // For each element of array.
    for (int i = 0; i < n; i++)
 
        // Try to find the negative value of
        // arr[i] from i + 1 to n
        for (int j = i + 1; j < n; j++)
 
            // If absolute values are equal print pair.
            if (abs(arr[i]) == abs(arr[j]))
                v.push_back(abs(arr[i]));     
 
    // If size of vector is 0, therefore there is no
    // element with positive negative value, print "0"
    if (v.size() == 0)
       return;
 
    // Sort the vector
    sort(v.begin(), v.end());
 
    // Print the pair with negative positive value.
    for (int i = 0; i < v.size(); i++)
        cout << -v[i] << " " << v[i];   
}
 
// Driven Program
int main()
{
    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printPairs(arr, n);
    return 0;
}

Java

// Java program to find pairs of positive
// and negative values present in an array.
import java.util.*;
import java.lang.*;
 
class GFG {
 
    // Print pair with negative and positive value
    public static void printPairs(int arr[] , int n)
    {
        Vector v = new Vector();
        // For each element of array.
        for (int i = 0; i < n; i++)
 
            // Try to find the negative value of
            // arr[i] from i + 1 to n
            for (int j = i + 1; j < n; j++)
 
                // If absolute values are equal
                // print pair.
                if (Math.abs(arr[i]) ==
                                  Math.abs(arr[j]))
                    v.add(Math.abs(arr[i]));
 
 
        // If size of vector is 0, therefore there
        // is no element with positive negative
        // value, print "0"
        if (v.size() == 0)
            return;    
     
        // Sort the vector
        Collections.sort(v);
 
        // Print the pair with negative positive
        // value.
        for (int i = 0; i < v.size(); i++)
            System.out.print(-v.get(i) + " " +
                                      v.get(i));
    }
 
    // Driven Program
    public static void main(String[] args)
    {
        int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
        int n = arr.length;
        printPairs(arr, n);
    }
}
 
// This code is contributed by Prasad Kshirsagar.

Python 3

# Simple Python 3 program to find
# pairs of positive and negative
# values present in an array.
 
# Print pair with negative and
# positive value
def printPairs(arr, n):
    v = []
 
    # For each element of array.
    for i in range(n):
 
        # Try to find the negative value
        # of arr[i] from i + 1 to n
        for j in range( i + 1,n) :
 
            # If absolute values are
            # equal print pair.
            if (abs(arr[i]) == abs(arr[j])) :
                v.append(abs(arr[i]))
 
    # If size of vector is 0, therefore
    # there is no element with positive
    # negative value, print "0"
    if (len(v) == 0):
        return;
 
    # Sort the vector
    v.sort()
 
    # Print the pair with negative
    # positive value.
    for i in range(len( v)):
        print(-v[i], "" , v[i], end = " ")
 
# Driver Code
if __name__ == "__main__":
    arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ]
    n = len(arr)
    printPairs(arr, n)
 
# This code is contributed
# by ChitraNayal

C#

// C# program to find pairs of positive
// and negative values present in an array.
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Print pair with negative and positive value
    public static void printPairs(int []arr , int n)
    {
        List v = new List();
         
        // For each element of array.
        for (int i = 0; i < n; i++)
 
            // Try to find the negative value of
            // arr[i] from i + 1 to n
            for (int j = i + 1; j < n; j++)
 
                // If absolute values are equal
                // print pair.
                if (Math.Abs(arr[i]) ==
                                Math.Abs(arr[j]))
                    v.Add(Math.Abs(arr[i]));
 
 
        // If size of vector is 0, therefore there
        // is no element with positive negative
        // value, print "0"
        if (v.Count == 0)
            return;    
     
        // Sort the vector
        v.Sort();
 
        // Print the pair with negative positive
        // value.
        for (int i = 0; i < v.Count; i++)
            Console.Write(-v[i] + " " +
                                    v[i]);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 4, 8, 9, -4, 1, -1, -8, -9 };
        int n = arr.Length;
        printPairs(arr, n);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

// Javascript program to find pairs of positive
// and negative values present in an array.
     
    // Print pair with negative and positive value
    function printPairs(arr,n)
    {
        let  v = [];
         
        // For each element of array.
        for (let i = 0; i < n; i++)
  
            // Try to find the negative value of
            // arr[i] from i + 1 to n
            for (let j = i + 1; j < n; j++)
  
                // If absolute values are equal
                // print pair.
                if (Math.abs(arr[i]) ==
                                  Math.abs(arr[j]))
                    v.push(Math.abs(arr[i]));
  
  
        // If size of vector is 0, therefore there
        // is no element with positive negative
        // value, print "0"
        if (v.length == 0)
            return;   
      
        // Sort the vector
        v.sort(function(a, b){return a-b;});
  
        // Print the pair with negative positive
        // value.
        for (let i = 0; i < v.length; i++)
            document.write(-v[i] + " " +
                                      v[i]);
    }
     
    // Driven Program
    let arr=[4, 8, 9, -4, 1, -1, -8, -9];
    let n = arr.length;
    printPairs(arr, n);
     
    // This code is contributed by rag2127

输出：

-1 1-4 4-8 8-9 9

方法二（散列）
这个想法是使用散列。遍历给定的数组，在哈希表的绝对值处增加计数。如果计数变为 2，则将其绝对值存储在另一个向量中。最后对向量进行排序。如果向量的大小为 0，则打印“0”，否则对于向量中的每一项，首先打印其负值和正值。

下面是这个方法的实现：

C++

// Efficient CPP program to find pairs of
// positive and negative values present in
// an array.
#include 
using namespace std;
 
// Print pair with negative and positive value
void printPairs(int arr[], int n)
{
    vector v;
    unordered_map cnt;
 
    // For each element of array.
    for (int i = 0; i < n; i++) {
 
        // If element has not encounter early,
        // mark it on cnt array.
        if (cnt[abs(arr[i])] == 0)
            cnt[abs(arr[i])] = 1;
 
        // If seen before, push it in vector (
        // given that elements are distinct)
        else {
            v.push_back(abs(arr[i]));
            cnt[abs(arr[i])] = 0;
        }
    }
 
    if (v.size() == 0)
        return;
 
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); i++)
        cout << -v[i] << " " << v[i] << " ";
}
 
// Driven Program
int main()
{
    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printPairs(arr, n);
    return 0;
}

Java

// Efficient Java program to find pairs of
// positive and negative values present in
// an array.
import java.util.*;
 
class GFG
{
 
// Print pair with negative
// and positive value
static void printPairs(int arr[], int n)
{
    ArrayList v = new ArrayList ();
    HashMap cnt = new HashMap();
 
    // For each element of array.
    for (int i = 0; i < n; i++)
    {
 
        // If element has  encounter early,
        // then increment its count
        if (cnt.containsKey(Math.abs(arr[i])))
            cnt.put(Math.abs(arr[i]) , cnt.get(Math.abs(arr[i])) + 1);
 
        // If element has not seen before,
          // then initialize its count to 1
        else
        {
              
            cnt.put(Math.abs(arr[i]), 1);
        }
        if( cnt.get(Math.abs(arr[i])) == 2 ){
          v.add(Math.abs(arr[i]));
        }
    }
 
    if (v.size() == 0)
        return;
 
    Collections.sort(v);
    for (int i = 0; i < v.size(); i++)
        System.out.print("-" + v.get(i) +
                         " " + v.get(i) + " ");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
    int n = arr.length;
    printPairs(arr, n);
}
}
 
// This code is contributed by Prerna Saini

蟒蛇3

# Efficient Python3 program to find pairs of 
# positive and negative values present in
# an array. 
 
# Print pair with negative and
# positive value
def printPairs(arr, n):
     
    s = set()
    ret = []
     
    # For each element of array.
    for i in arr:
        if abs(i) in s:
            ret.append(abs(i))
        else:
            s.add(abs(i))
             
    ret.sort()
     
    for i in range(0, len(ret)):
        print(-ret[i], "", ret[i], end = " ")
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ]
    n = len(arr)
     
    printPairs(arr, n)
 
# This code is contributed by RohitOberoi

C＃

// Efficient C# program to find pairs of
// positive and negative values present in
// an array.
using System;
using System.Collections.Generic;
class GFG
{
     
    // Print pair with negative and positive value
    static void printPairs(int[] arr, int n)
    {
        List v = new List();
        Dictionary cnt = new Dictionary();
      
        // For each element of array.
        for (int i = 0; i < n; i++)
        {
      
            // If element has not encounter early,
            // mark it on cnt array.
            if(!cnt.ContainsKey(Math.Abs(arr[i])))
            {
                cnt[Math.Abs(arr[i])] = true;
            }
            else if(cnt[Math.Abs(arr[i])] == false)
            {
                cnt[Math.Abs(arr[i])] = true;
            }
            else
            {
                v.Add(Math.Abs(arr[i]));
                cnt[Math.Abs(arr[i])] = false;
            }
        }
      
        if (v.Count == 0)
            return;
      
        v.Sort();
        for (int i = 0; i < v.Count; i++)
            Console.Write(-v[i] + " " + v[i] + " ");
    }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };
    int n = arr.Length;
    printPairs(arr, n);
  }
}
 
// This code is contributed by divyeshrabadiya07

输出：

-1 1 -4 4 -8 8 -9 9

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