算法设计：找到所有零和的三元组

2021年3月15日09:42:22 发表评论 902 次浏览

本文概述

给定一系列不同的元素。任务是在总和为零的数组中查找三元组。

例子 ：

Input : arr[] = {0, -1, 2, -3, 1}
Output : (0 -1 1), (2 -3 1)

Explanation : The triplets with zero sum are
0 + -1 + 1 = 0 and 2 + -3 + 1 = 0  

Input : arr[] = {1, -2, 1, 0, 5}
Output : 1 -2  1
Explanation : The triplets with zero sum is
1 + -2 + 1 = 0

推荐：请在"实践首先, 在继续解决方案之前。

方法1：这是一个简单的方法, 需要O(n3)的时间得出结果。

方法：

天真的方法会运行三个循环, 并一一检查三个元素的总和是否为零。如果三个元素的总和为零, 则打印元素, 否则找不到打印。

算法：

  1. 使用循环计数器运行三个嵌套循环一世, j, k
  2. 这三个循环从0到n-3, 第二个循环从i + 1到n-2, 第三个循环从j + 1到n-1。循环计数器代表三元组的三个元素。
  3. 检查第i, 第j, 第k个元素的总和是否等于零。如果是, 则打印总和, 否则继续。

实现

C ++

// A simple C++ program to find three elements
// whose sum is equal to zero
#include<bits/stdc++.h>
using namespace std;
  
// Prints all triplets in arr[] with 0 sum
void findTriplets( int arr[], int n)
{
     bool found = true ;
     for ( int i=0; i<n-2; i++)
     {
         for ( int j=i+1; j<n-1; j++)
         {
             for ( int k=j+1; k<n; k++)
             {
                 if (arr[i]+arr[j]+arr[k] == 0)
                 {
                     cout << arr[i] << " "
                          << arr[j] << " "
                          << arr[k] <<endl;
                     found = true ;
                 }
             }
         }
     }
  
     // If no triplet with 0 sum found in array
     if (found == false )
         cout << " not exist " <<endl;
  
}
  
// Driver code
int main()
{
     int arr[] = {0, -1, 2, -3, 1};
     int n = sizeof (arr)/ sizeof (arr[0]);
     findTriplets(arr, n);
     return 0;
}

Java

// A simple Java program to find three elements
// whose sum is equal to zero
class num{
// Prints all triplets in arr[] with 0 sum
static void findTriplets( int [] arr, int n)
{
     boolean found = true ;
     for ( int i= 0 ; i<n- 2 ; i++)
     {
         for ( int j=i+ 1 ; j<n- 1 ; j++)
         {
             for ( int k=j+ 1 ; k<n; k++)
             {
                 if (arr[i]+arr[j]+arr[k] == 0 )
                 {
                     System.out.print(arr[i]);
                     System.out.print( " " );
                     System.out.print(arr[j]);
                     System.out.print( " " );
                     System.out.print(arr[k]);
                     System.out.print( "\n" );
                     found = true ;
                 }
             }
         }
     }
  
     // If no triplet with 0 sum found in array
     if (found == false )
         System.out.println( " not exist " );
  
}
  
// Driver code
public static void main(String[] args)
{
     int arr[] = { 0 , - 1 , 2 , - 3 , 1 };
     int n =arr.length;
     findTriplets(arr, n);
  
}
}
//This code is contributed by
//Smitha Dinesh Semwal

Python3

# A simple Python 3 program 
# to find three elements whose 
# sum is equal to zero
  
# Prints all triplets in 
# arr[] with 0 sum
def findTriplets(arr, n):
  
     found = True
     for i in range ( 0 , n - 2 ):
      
         for j in range (i + 1 , n - 1 ):
          
             for k in range (j + 1 , n):
              
                 if (arr[i] + arr[j] + arr[k] = = 0 ):
                     print (arr[i], arr[j], arr[k])
                     found = True
      
              
     # If no triplet with 0 sum 
     # found in array
     if (found = = False ):
         print ( " not exist " )
  
# Driver code
arr = [ 0 , - 1 , 2 , - 3 , 1 ]
n = len (arr)
findTriplets(arr, n)
  
# This code is contributed by Smitha Dinesh Semwal

C#

// A simple C# program to find three elements 
// whose sum is equal to zero
using System;
  
class GFG {
      
     // Prints all triplets in arr[] with 0 sum
     static void findTriplets( int []arr, int n)
     {
         bool found = true ;
         for ( int i = 0; i < n-2; i++)
         {
             for ( int j = i+1; j < n-1; j++)
             {
                 for ( int k = j+1; k < n; k++)
                 {
                     if (arr[i] + arr[j] + arr[k]
                                            == 0)
                     {
                         Console.Write(arr[i]);
                         Console.Write( " " );
                         Console.Write(arr[j]);
                         Console.Write( " " );
                         Console.Write(arr[k]);
                         Console.Write( "\n" );
                         found = true ;
                     }
                 }
             }
         }
      
         // If no triplet with 0 sum found in 
         // array
         if (found == false )
             Console.Write( " not exist " );
     }
      
     // Driver code
     public static void Main()
     {
         int []arr = {0, -1, 2, -3, 1};
         int n = arr.Length;
         findTriplets(arr, n);
     }
}
  
// This code is contributed by nitin mittal.

的PHP

<?php
// A simple PHP program to 
// find three elements whose 
// sum is equal to zero
  
// Prints all triplets
// in arr[] with 0 sum
function findTriplets( $arr , $n )
{
     $found = true;
     for ( $i = 0; $i < $n - 2; $i ++)
     {
         for ( $j = $i + 1; $j < $n - 1; $j ++)
         {
             for ( $k = $j + 1; $k < $n ; $k ++)
             {
                 if ( $arr [ $i ] + $arr [ $j ] + 
                                $arr [ $k ] == 0)
                 {
                     echo $arr [ $i ] , " " , $arr [ $j ] , " " , $arr [ $k ] , "\n" ;
                     $found = true;
                 }
             }
         }
     }
  
     // If no triplet with 0
     // sum found in array
     if ( $found == false)
         echo " not exist " , "\n" ;
  
}
  
// Driver Code
$arr = array (0, -1, 2, -3, 1);
$n = sizeof( $arr );
findTriplets( $arr , $n );
  
// This code is contributed by m_kit
?>

输出如下：

0 -1 1
2 -3 1

复杂度分析：

  • 时间复杂度：上3)。
    由于需要三个嵌套循环, 因此时间复杂度为O(n3)。
  • 辅助空间：O(1)。
    由于不需要额外的空间, 因此时间复杂度是恒定的。

方法2：第二种方法使用散列处理得出结果, 并在较短的O(n2)。

方法：

这涉及遍历数组。对于每个元素arr [i], 找到一对总和为" -arr [i]"的对。这个问题简化为成对和, 可以使用哈希在O(n)时间内解决。

算法：

  1. 创建一个哈希来存储键值对。
  2. 运行带有两个循环的嵌套循环, 外循环从0到n-2, 内循环从i + 1到n-1
  3. 检查哈希图中是否存在第ith个元素和第j个元素的和与-1相乘
  4. 如果该元素存在于哈希图中, 则打印三元组, 否则将第j个元素插入哈希图中。

实现

C ++

// C++ program to find triplets in a given
// array whose sum is zero
#include<bits/stdc++.h>
using namespace std;
  
// function to print triplets with 0 sum
void findTriplets( int arr[], int n)
{
     bool found = false ;
  
     for ( int i=0; i<n-1; i++)
     {
         // Find all pairs with sum equals to
         // "-arr[i]"
         unordered_set< int > s;
         for ( int j=i+1; j<n; j++)
         {
             int x = -(arr[i] + arr[j]);
             if (s.find(x) != s.end())
             {
                 printf ( "%d %d %d\n" , x, arr[i], arr[j]);
                 found = true ;
             }
             else
                 s.insert(arr[j]);
         }
     }
  
     if (found == false )
         cout << " No Triplet Found" << endl;
}
  
// Driver code
int main()
{
     int arr[] = {0, -1, 2, -3, 1};
     int n = sizeof (arr)/ sizeof (arr[0]);
     findTriplets(arr, n);
     return 0;
}

Java

// Java program to find triplets in a given
// array whose sum is zero
import java.util.*;
  
class GFG 
{
  
     // function to print triplets with 0 sum
     static void findTriplets( int arr[], int n) 
     {
         boolean found = false ;
  
         for ( int i = 0 ; i < n - 1 ; i++) 
         {
             // Find all pairs with sum equals to
             // "-arr[i]"
             HashSet<Integer> s = new HashSet<Integer>();
             for ( int j = i + 1 ; j < n; j++) 
             {
                 int x = -(arr[i] + arr[j]);
                 if (s.contains(x)) 
                 {
                     System.out.printf( "%d %d %d\n" , x, arr[i], arr[j]);
                     found = true ;
                 } 
                 else 
                 {
                     s.add(arr[j]);
                 }
             }
         }
  
         if (found == false )
         {
             System.out.printf( " No Triplet Found\n" );
         }
     }
  
     // Driver code
     public static void main(String[] args) 
     {
         int arr[] = { 0 , - 1 , 2 , - 3 , 1 };
         int n = arr.length;
         findTriplets(arr, n);
     }
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 program to find triplets 
# in a given array whose sum is zero 
  
# function to print triplets with 0 sum 
def findTriplets(arr, n):
     found = False
     for i in range (n - 1 ):
  
         # Find all pairs with sum 
         # equals to "-arr[i]" 
         s = set ()
         for j in range (i + 1 , n):
             x = - (arr[i] + arr[j])
             if x in s:
                 print (x, arr[i], arr[j])
                 found = True
             else :
                 s.add(arr[j])
     if found = = False :
         print ( "No Triplet Found" )
  
# Driver Code
arr = [ 0 , - 1 , 2 , - 3 , 1 ]
n = len (arr)
findTriplets(arr, n)
  
# This code is contributed by Shrikant13

C#

// C# program to find triplets in a given
// array whose sum is zero
using System;
using System.Collections.Generic;
  
class GFG 
{
  
     // function to print triplets with 0 sum
     static void findTriplets( int []arr, int n) 
     {
         bool found = false ;
  
         for ( int i = 0; i < n - 1; i++) 
         {
             // Find all pairs with sum equals to
             // "-arr[i]"
             HashSet< int > s = new HashSet< int >();
             for ( int j = i + 1; j < n; j++) 
             {
                 int x = -(arr[i] + arr[j]);
                 if (s.Contains(x)) 
                 {
                     Console.Write( "{0} {1} {2}\n" , x, arr[i], arr[j]);
                     found = true ;
                 } 
                 else
                 {
                     s.Add(arr[j]);
                 }
             }
         }
  
         if (found == false )
         {
             Console.Write( " No Triplet Found\n" );
         }
     }
  
     // Driver code
     public static void Main(String[] args) 
     {
         int []arr = {0, -1, 2, -3, 1};
         int n = arr.Length;
         findTriplets(arr, n);
     }
}
  
// This code has been contributed by 29AjayKumar

输出如下：

-1 0 1
-3 2 1

复杂度分析：

  • 时间复杂度：上2)。
    由于需要两个嵌套循环, 因此时间复杂度为O(n2)。
  • 辅助空间：上)。
    由于需要一个哈希图, 因此时间复杂度是线性的。

方法3：该方法使用排序来得出正确的结果, 并在O(n2) 时间。

方法：

上述方法需要额外的空间。这个想法是基于方法2

这个

发布。对于每个元素, 检查是否存在一对总和等于该元素的负值的对。

算法：

  1. 以升序对数组进行排序。
  2. 从头到尾遍历数组。
  3. 对于每个索引一世, 创建两个变量l =我+ 1和r = n – 1
  4. 循环运行直到l小于r, 如果array [l], array [r]的总和等于零, 则打印三元组并中断循环
  5. 如果总和小于零, 则增加l的值, 通过增加l的值, 总和将随着数组的排序而增加, 因此数组[l + 1]>数组[l]
  6. 如果总和大于零, 那么递减r的值, 通过增加l的值, 总和将随着数组的排序而减少, 因此array [r-1] <数组[r].

实现

C ++

// C++ program to find triplets in a given
// array whose sum is zero
#include<bits/stdc++.h>
using namespace std;
  
// function to print triplets with 0 sum
void findTriplets( int arr[], int n)
{
     bool found = false ;
  
     // sort array elements
     sort(arr, arr+n);
  
     for ( int i=0; i<n-1; i++)
     {
         // initialize left and right
         int l = i + 1;
         int r = n - 1;
         int x = arr[i];
         while (l < r)
         {
             if (x + arr[l] + arr[r] == 0)
             {
                 // print elements if it's sum is zero
                 printf ( "%d %d %d\n" , x, arr[l], arr[r]);
                 l++;
                 r--;
                 found = true ;
             }
  
             // If sum of three elements is less
             // than zero then increment in left
             else if (x + arr[l] + arr[r] < 0)
                 l++;
  
             // if sum is greater than zero than
             // decrement in right side
             else
                 r--;
         }
     }
  
     if (found == false )
         cout << " No Triplet Found" << endl;
}
  
// Driven source
int main()
{
     int arr[] = {0, -1, 2, -3, 1};
     int n = sizeof (arr)/ sizeof (arr[0]);
     findTriplets(arr, n);
     return 0;
}

Java

// Java  program to find triplets in a given
// array whose sum is zero
import java.util.Arrays; 
import java.io.*;
  
class GFG {
     // function to print triplets with 0 sum
static void findTriplets( int arr[], int n)
{
     boolean found = false ;
  
     // sort array elements
     Arrays.sort(arr);
  
     for ( int i= 0 ; i<n- 1 ; i++)
     {
         // initialize left and right
         int l = i + 1 ;
         int r = n - 1 ;
         int x = arr[i];
         while (l < r)
         {
             if (x + arr[l] + arr[r] == 0 )
             {
                 // print elements if it's sum is zero
                     System.out.print(x + " " );
                     System.out.print(arr[l]+ " " );
                     System.out.println(arr[r]+ " " );
      
                 l++;
                 r--;
                 found = true ;
             }
  
             // If sum of three elements is less
             // than zero then increment in left
             else if (x + arr[l] + arr[r] < 0 )
                 l++;
  
             // if sum is greater than zero than
             // decrement in right side
             else
                 r--;
         }
     }
  
     if (found == false )
             System.out.println( " No Triplet Found" );
}
  
// Driven source
     public static void main (String[] args) {
  
     int arr[] = { 0 , - 1 , 2 , - 3 , 1 };
     int n =arr.length;
     findTriplets(arr, n);
     }
//This code is contributed by Tushil..    
}

Python3

# python program to find triplets in a given
# array whose sum is zero
  
# function to print triplets with 0 sum
def findTriplets(arr, n):
  
     found = False
  
     # sort array elements
     arr.sort()
  
     for i in range ( 0 , n - 1 ):
      
         # initialize left and right
         l = i + 1
         r = n - 1
         x = arr[i]
         while (l < r):
          
             if (x + arr[l] + arr[r] = = 0 ):
                 # print elements if it's sum is zero
                 print (x, arr[l], arr[r])
                 l + = 1
                 r - = 1
                 found = True
              
  
             # If sum of three elements is less
             # than zero then increment in left
             elif (x + arr[l] + arr[r] < 0 ):
                 l + = 1
  
             # if sum is greater than zero than
             # decrement in right side
             else :
                 r - = 1
          
     if (found = = False ):
         print ( " No Triplet Found" )
  
  
# Driven source
arr = [ 0 , - 1 , 2 , - 3 , 1 ]
n = len (arr)
findTriplets(arr, n)
  
# This code is contributed by Smitha Dinesh Semwal

C#

// C#  program to find triplets in a given
// array whose sum is zero
using System;
  
public class GFG{
         // function to print triplets with 0 sum
static void findTriplets( int []arr, int n)
{
     bool found = false ;
  
     // sort array elements
     Array.Sort(arr);
  
     for ( int i=0; i<n-1; i++)
     {
         // initialize left and right
         int l = i + 1;
         int r = n - 1;
         int x = arr[i];
         while (l < r)
         {
             if (x + arr[l] + arr[r] == 0)
             {
                 // print elements if it's sum is zero
                     Console.Write(x + " " );
                     Console.Write(arr[l]+ " " );
                     Console.WriteLine(arr[r]+ " " );
      
                 l++;
                 r--;
                 found = true ;
             }
  
             // If sum of three elements is less
             // than zero then increment in left
             else if (x + arr[l] + arr[r] < 0)
                 l++;
  
             // if sum is greater than zero than
             // decrement in right side
             else
                 r--;
         }
     }
  
     if (found == false )
             Console.WriteLine( " No Triplet Found" );
}
  
// Driven source
     static public void Main (){
          
     int []arr = {0, -1, 2, -3, 1};
     int n =arr.Length;
     findTriplets(arr, n);
     }
//This code is contributed by akt_mit.. 
}

的PHP

<?php
// PHP program to find 
// triplets in a given
// array whose sum is zero
  
// function to print 
// triplets with 0 sum
function findTriplets( $arr , $n )
{
     $found = false;
  
     // sort array elements
     sort( $arr );
  
     for ( $i = 0; $i < $n - 1; $i ++)
     {
         // initialize left
         // and right
         $l = $i + 1;
         $r = $n - 1;
         $x = $arr [ $i ];
         while ( $l < $r )
         {
             if ( $x + $arr [ $l ] + 
                      $arr [ $r ] == 0)
             {
                 // print elements if 
                 // it's sum is zero
                 echo $x , " " , $arr [ $l ], " " , $arr [ $r ], "\n" ;
                 $l ++;
                 $r --;
                 $found = true;
             }
  
             // If sum of three elements 
             // is less than zero then 
             // increment in left
             else if ( $x + $arr [ $l ] + 
                           $arr [ $r ] < 0)
                 $l ++;
  
             // if sum is greater than 
             // zero than decrement
             // in right side
             else
                 $r --;
         }
     }
  
     if ( $found == false)
         echo " No Triplet Found" , "\n" ;
}
  
// Driver Code
$arr = array (0, -1, 2, -3, 1);
$n = sizeof( $arr );
findTriplets( $arr , $n );
  
// This code is contributed by ajit
?>

输出：

-3 1 2
-1 0 1

复杂度分析：

  • 时间复杂度：上2)。
    只需要两个嵌套循环, 因此时间复杂度为O(n2)。
  • 辅助空间：O(1), 不需要额外的空间, 因此时间复杂度是恒定的。

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