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 = 0Input : arr[] = {1, -2, 1, 0, 5} Output : 1 -21 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。
【算法设计（找到所有零和的三元组）】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

// Javaprogram 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 - = 1if (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。

如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

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