在数组中查找总和已存在于数组中的对

给定一个由 n 个不同的正元素组成的数组，任务是找到其总和已经存在于给定数组中的对。
例子 ：

Input : arr[] = {2, 8, 7, 1, 5};
Output : 2 5
         7 1    
     
Input : arr[] = {7, 8, 5, 9, 11};
Output : Not Exist

一种朴素的方法是运行三个循环来找到其总和存在于数组中的对。

C++

// A simple C++ program to find pair whose sum
// already exists in array
#include 
using namespace std;
 
// Function to find pair whose sum exists in arr[]
void findPair(int arr[], int n)
{
    bool found = false;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = 0; k < n; k++) {
                if (arr[i] + arr[j] == arr[k]) {
                    cout << arr[i] << " " << arr[j] << endl;
                    found = true;
                }
            }
        }
    }
 
    if (found == false)
        cout << "Not exist" << endl;
}
 
// Driven code
int main()
{
    int arr[] = { 10, 4, 8, 13, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findPair(arr, n);
    return 0;
}

Java

// A simple Java program to find
// pair whose sum already exists
// in array
import java.io.*;
 
public class GFG {
 
    // Function to find pair whose
    // sum exists in arr[]
    static void findPair(int[] arr, int n)
    {
        boolean found = false;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        System.out.println(arr[i] +
                                      " " + arr[j]);
                        found = true;
                    }
                }
            }
        }
 
        if (found == false)
            System.out.println("Not exist");
    }
 
    // Driver code
    static public void main(String[] args)
    {
        int[] arr = {10, 4, 8, 13, 5};
        int n = arr.length;
        findPair(arr, n);
    }
}
 
// This code is contributed by vt_m.

Python3

# A simple python program to find pair
# whose sum already exists in array
 
# Function to find pair whose sum
# exists in arr[]
def findPair(arr, n):
    found = False
    for i in range(0, n):
        for j in range(i + 1, n):
            for k in range(0, n):
                if (arr[i] + arr[j] == arr[k]):
                    print(arr[i], arr[j])
                    found = True
 
    if (found == False):
        print("Not exist")
 
# Driver code
if __name__ == '__main__':
    arr = [ 10, 4, 8, 13, 5 ]
    n = len(arr)
    findPair(arr, n)
     
# This code contributed by 29AjayKumar

C#

// A simple C# program to find
// pair whose sum already exists
// in array
using System;
 
public class GFG {
 
    // Function to find pair whose
    // sum exists in arr[]
    static void findPair(int[] arr, int n)
    {
        bool found = false;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        Console.WriteLine(arr[i] +
                                      " " + arr[j]);
                        found = true;
                    }
                }
            }
        }
 
        if (found == false)
            Console.WriteLine("Not exist");
    }
 
    // Driver code
    static public void Main(String []args)
    {
        int[] arr = {10, 4, 8, 13, 5};
        int n = arr.Length;
        findPair(arr, n);
    }
}
 
// This code is contributed by vt_m.

Javascript

// A simple Javascript program to find
// pair whose sum already exists
// in array
     
    // Function to find pair whose
    // sum exists in arr[]
    function findPair(arr,n)
    {
        let found = false;
        for (let i = 0; i < n; i++) {
            for (let j = i + 1; j < n; j++) {
                for (let k = 0; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        document.write(arr[i] +
                                      " " + arr[j]+"<br>");
                        found = true;
                    }
                }
            }
        }
   
        if (found == false)
            document.write("Not exist");
    }
     
    // Driver code
    let arr=[10, 4, 8, 13, 5];
    let n = arr.length;
    findPair(arr, n);
     
     
 
// This code is contributed by patel2127

输出 ：

8 5

一个有效的解决方案是将所有元素存储在一个哈希表中（在 C++ 中为 unordered_set）并一一检查所有对并检查其总和是否存在于集合中。如果它存在于集合中，则打印对。如果在数组中没有找到对，则打印不存在。

C++

// C++ program to find pair whose sum already
// exists in array
#include 
using namespace std;
 
// Function to find pair whose sum exists in arr[]
void findPair(int arr[], int n)
{
    // Hash to store all element of array
    unordered_set s;
    for (int i = 0; i < n; i++)
        s.insert(arr[i]);
 
    bool found = false;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // Check sum already exists or not
            if (s.find(arr[i] + arr[j]) != s.end()) {
                cout << arr[i] << " " << arr[j] << endl;
                found = true;
            }
        }
    }
 
    if (found == false)
        cout << "Not exist" << endl;
}
 
// Driven code
int main()
{
    int arr[] = { 10, 4, 8, 13, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findPair(arr, n);
    return 0;
}

Java

// Java program to find pair whose sum
// already exists in array
import java.util.*;
import java.lang.*;
import java.io.*;
 
class Getpairs {
    // Function to find pair whose sum
    // exists in arr[]
    public static void findPair(int[] arr, int n)
    {
        /* store all the array elements as a
        Hash value*/
        HashSet s = new HashSet();
 
        for (Integer i : arr) {
            s.add(i);
        }
 
        /* Run two loop and check for the sum
    in the Hashset*/
        /* if not a single pair exist then found
    will be false else true*/
        boolean found = false;
 
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = arr[i] + arr[j];
                if (s.contains(sum)) {
                    /* if the sum is present in
                 hashset then found become
                true*/
                    found = true;
 
                    System.out.println(arr[i] + " "
                                       + arr[j]);
                }
            }
        }
 
        if (found == false)
            System.out.println("Not Exist ");
    }
 
    // driver function
    public static void main(String args[])
    {
        int[] arr = { 10, 4, 8, 13, 5 };
        int n = arr.length;
        findPair(arr, n);
    }
}
 
// This code is contributed by Smarak Chopdar

蟒蛇3

# Python3 program to find pair whose
# sum already exist in arrar
 
# Function to find pair whose
# sum sxists in arr[]
def findPair(arr, n):
     
    # hash to store all element of array
    s = {i : 1 for i in arr}
     
    found = False
     
    for i in range(n):
        for j in range(i + 1, n):
             
            # check if sum already exists or not
            if arr[i] + arr[j] in s.keys():
                print(arr[i], arr[j])
                found = True
    if found == False:
        print("Not exist")
         
# Driver code
arr = [10, 4, 8, 13, 5]
 
n = len(arr)
 
findPair(arr, n)
     
# This code is contributed
# by Mohit Kumar

C＃

// C# program to find pair whose sum
// already exists in array
using System;
using System.Collections.Generic;
 
class Getpairs
{
    // Function to find pair whose sum
    // exists in arr[]
    public static void findPair(int[] arr, int n)
    {
        /* store all the array elements as a
        Hash value*/
        HashSet s = new HashSet();
 
        foreach (int i in arr)
        {
            s.Add(i);
        }
 
        /* Run two loop and check for the sum
    in the Hashset*/
        /* if not a single pair exist then found
    will be false else true*/
        bool found = false;
 
        for (int i = 0; i < n - 1; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                int sum = arr[i] + arr[j];
                if (s.Contains(sum))
                {
                    /* if the sum is present in
                    hashset then found become
                    true*/
                    found = true;
 
                    Console.WriteLine(arr[i] + " "
                                    + arr[j]);
                }
            }
        }
 
        if (found == false)
            Console.WriteLine("Not Exist ");
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 10, 4, 8, 13, 5 };
        int n = arr.Length;
        findPair(arr, n);
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript program to find pair whose sum
// already exists in array   
     
    // Function to find pair whose sum
    // exists in arr[]
    function findPair(arr,n)
    {
        /* store all the array elements as a
        Hash value*/
        let s = new Set();
  
        for (let i=0;i<arr.length;i++) {
            s.add(arr[i]);
        }
  
        /* Run two loop and check for the sum
    in the Hashset*/
        /* if not a single pair exist then found
    will be false else true*/
        let found = false;
  
        for (let i = 0; i < n - 1; i++) {
            for (let j = i + 1; j < n; j++) {
                let sum = arr[i] + arr[j];
                if (s.has(sum)) {
                    /* if the sum is present in
                 hashset then found become
                true*/
                    found = true;
  
                    document.write(arr[i] + " "
                                  + arr[j]+"<br>");
                }
            }
        }
  
        if (found == false)
            document.write("Not Exist ");
    }
     
    // driver function
    let arr=[10, 4, 8, 13, 5 ];
    let n = arr.length;
    findPair(arr, n);
     
// This code is contributed by unknown2108
 

输出：

8 5

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