打印具有给定总和的所有偶对
原文：https://www.geeksforgeeks.org/print-all-pairs-with-given-sum/
给定一个整数数组和一个数字sum，请打印该数组中所有总和等于sum的对。
Examples :
Input : arr[] = {1, 5, 7, -1, 5},
sum = 6
Output : (1, 5) (7, -1) (1, 5)
Input : arr[] = {2, 5, 17, -1},
sum = 7
Output : (2, 5)
一种简单的解决方案是遍历每个元素，并检查数组中是否存在可以添加到其上的另一个数字以求和。
C++
// C++ implementation of simple method to
// find print pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
int printPairs(int arr[], int n, int sum)
{
int count = 0; // Initialize result
// Consider all possible pairs and check
// their sums
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] + arr[j] == sum)
cout << "(" << arr[i] << ", "
<< arr[j] << ")" << endl;
}
// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 6;
printPairs(arr, n, sum);
return 0;
}
Java
// Java implementation of
// simple method to find
// print pairs with given sum.
class GFG {
// Returns number of pairs
// in arr[0..n-1] with sum
// equal to 'sum'
static void printPairs(int arr[],
int n, int sum)
{
// int count = 0;
// Consider all possible pairs
// and check their sums
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] + arr[j] == sum)
System.out.println("(" + arr[i] + ", " + arr[j] + ")");
}
// Driver Code
public static void main(String[] arg)
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = arr.length;
int sum = 6;
printPairs(arr, n, sum);
}
}
// This code is contributed
// by Smitha
Python3
# Python 3 implementation
# of simple method to find
# print pairs with given sum.
# Returns number of pairs
# in arr[0..n-1] with sum
# equal to 'sum'
def printPairs(arr, n, sum):
# count = 0
# Consider all possible
# pairs and check their sums
for i in range(0, n ):
for j in range(i + 1, n ):
if (arr[i] + arr[j] == sum):
print("(", arr[i],
", ", arr[j],
")", sep = "")
# Driver Code
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
printPairs(arr, n, sum)
# This code is contributed
# by Smitha
C
// C# implementation of simple
// method to find print pairs
// with given sum.
using System;
class GFG {
// Returns number of pairs
// in arr[0..n-1] with sum
// equal to 'sum'
static void printPairs(int[] arr,
int n, int sum)
{
// int count = 0;
// Consider all possible pairs
// and check their sums
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] + arr[j] == sum)
Console.Write("(" + arr[i] + ", " + arr[j] + ")"
+ "\n");
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 5, 7, -1, 5 };
int n = arr.Length;
int sum = 6;
printPairs(arr, n, sum);
}
}
// This code is contributed
// by Smitha
PHP
<?php
// PHP implementation of simple
// method to find print pairs
// with given sum.
// Returns number of pairs in
// arr[0..n-1] with sum equal
// to 'sum'
function printPairs($arr, $n, $sum)
{
// Initialize result
$count = 0;
// Consider all possible
// pairs and check their sums
for ($i = 0; $i < $n; $i++)
for ( $j = $i + 1; $j < $n; $j++)
if ($arr[$i] + $arr[$j] == $sum)
echo "(", $arr[$i], ", ",
$arr[$j], ")", "\n";
}
// Driver Code
$arr = array (1, 5, 7, -1, 5);
$n = sizeof($arr);
$sum = 6;
printPairs($arr, $n, $sum);
// This code is contributed by m_kit
?>
输出：
(1, 5)
(1, 5)
(7, -1)
方法 2（使用哈希）。
我们创建一个空的哈希表。 现在，我们遍历数组并检查哈希表中的对。 如果找到匹配的元素，我们将打印出等于该匹配元素出现次数的对数。
请注意，此解决方案最复杂的时间复杂度是O(c + n)，其中c是具有给定总和的对的计数。
C++
// C++ implementation of simple method to
// find count of pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
void printPairs(int arr[], int n, int sum)
{
// Store counts of all elements in map m
unordered_map<int, int> m;
// Traverse through all elements
for (int i = 0; i < n; i++) {
// Search if a pair can be formed with
// arr[i].
int rem = sum - arr[i];
if (m.find(rem) != m.end()) {
int count = m[rem];
for (int j = 0; j < count; j++)
cout << "(" << rem << ", "
<< arr[i] << ")" << endl;
}
m[arr[i]]++;
}
}
// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 6;
printPairs(arr, n, sum);
return 0;
}
Java
// Java implementation of simple method to
// find count of pairs with given sum.
import java.util.*;
class GFG{
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
static void printPairs(int arr[], int n,
int sum)
{
// Store counts of all elements in map m
HashMap<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// Traverse through all elements
for(int i = 0; i < n; i++)
{
// Search if a pair can be formed with
// arr[i].
int rem = sum - arr[i];
if (mp.containsKey(rem))
{
int count = mp.get(rem);
for(int j = 0; j < count; j++)
System.out.print("(" + rem +
", " + arr[i] +
")" + "\n");
}
if (mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.put(arr[i], 1);
}
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = arr.length;
int sum = 6;
printPairs(arr, n, sum);
}
}
// This code is contributed by Princi Singh
C
// C# implementation of simple method to
// find count of pairs with given sum.
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
static void printPairs(int []arr, int n, int sum)
{
// Store counts of all elements in map m
Dictionary<int,
int> m = new Dictionary<int,
int>();
// Traverse through all elements
for(int i = 0; i < n; i++)
{
// Search if a pair can be formed with
// arr[i].
int rem = sum - arr[i];
if (m.ContainsKey(rem))
{
int count = m[rem];
for(int j = 0; j < count; j++)
{
Console.Write("(" + rem + ", " +
arr[i] + ")" + "\n");
}
}
if (m.ContainsKey(arr[i]))
{
m[arr[i]]++;
}
else
{
m[arr[i]] = 1;
}
}
}
// Driver code
public static void Main(string[] args)
{
int []arr = { 1, 5, 7, -1, 5 };
int n = arr.Length;
int sum = 6;
printPairs(arr, n, sum);
}
}
// This code is contributed by rutvik_56
输出：
(1, 5)
(7, -1)
(1, 5)
方法 3 。
另一种打印具有给定总和的对的方法如下：
C++
// C++ code to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
void pairedElements(int arr[],
int sum, int n)
{
int low = 0;
int high = n - 1;
while (low < high)
{
if (arr[low] + arr[high] == sum)
{
cout << "The pair is : (" <<
arr[low] << ", " <<
arr[high] << ")" << endl;
}
if (arr[low] + arr[high] > sum)
{
high--;
}
else
{
low++;
}
}
}
// Driver code
int main()
{
int arr[] = {2, 3, 4, -2,
6, 8, 9, 11};
int n = sizeof(arr) / sizeof(arr[0]);
sort(arr, arr + n);
pairedElements(arr, 6, n);
}
// This code is contributed by Rajput-Ji
Java
import java.util.Arrays;
/**
* Created by sampat.
*/
public class SumOfPairs {
public void pairedElements(int arr[], int sum)
{
int low = 0;
int high = arr.length - 1;
while (low < high) {
if (arr[low] + arr[high] == sum) {
System.out.println("The pair is : ("
+ arr[low] + ", " + arr[high] + ")");
}
if (arr[low] + arr[high] > sum) {
high--;
}
else {
low++;
}
}
}
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, -2, 6, 8, 9, 11 };
Arrays.sort(arr);
SumOfPairs sp = new SumOfPairs();
sp.pairedElements(arr, 6);
}
}
Python3
# Python3 program for the
# above approach
def pairedElements(arr, sum):
low = 0;
high = len(arr) - 1;
while (low < high):
if (arr[low] +
arr[high] == sum):
print("The pair is : (", arr[low],
", ", arr[high], ")");
if (arr[low] + arr[high] > sum):
high -= 1;
else:
low += 1;
# Driver code
if __name__ == '__main__':
arr = [2, 3, 4, -2,
6, 8, 9, 11];
arr.sort();
pairedElements(arr, 6);
# This code contributed by shikhasingrajput
C
// C# program to find triplets in a given
// array whose sum is equal to given sum.
using System;
public class SumOfPairs
{
public void pairedElements(int []arr, int sum)
{
int low = 0;
int high = arr.Length - 1;
while (low < high)
{
if (arr[low] + arr[high] == sum)
{
Console.WriteLine("The pair is : ("
+ arr[low] + ", " + arr[high] + ")");
}
if (arr[low] + arr[high] > sum)
{
high--;
}
else
{
low++;
}
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 3, 4, -2, 6, 8, 9, 11 };
Array.Sort(arr);
SumOfPairs sp = new SumOfPairs();
sp.pairedElements(arr, 6);
}
}
// This code is contributed by Princi Singh
输出：
The pair is : (-2, 8)
The pair is : (2, 4)
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