查找一个三元组，将其总和成给定值

给定一个数组和一个值，查找数组中是否有一个三元组，其总和等于给定值。如果数组中存在这样的三元组，则打印该三元组并返回true。否则返回false。
例子：

Input: array = {12, 3, 4, 1, 6, 9}, sum = 24;
Output: 12, 3, 9
Explanation: There is a triplet (12, 3 and 9) present
in the array whose sum is 24. 

Input: array = {1, 2, 3, 4, 5}, sum = 9
Output: 5, 3, 1
Explanation: There is a triplet (5, 3 and 1) present 
in the array whose sum is 9. 

方法1 ：这是解决上述问题的幼稚方法。

• 方法：一种简单的方法是生成所有可能的三元组，并将每个三元组的总和与给定值进行比较。以下代码使用三个嵌套循环实现了此简单方法。
• 算法：
  1. 给定一个长度为n和总和为s的数组
  2. 创建三个嵌套循环，第一个循环从头到尾运行(循环计数器i)，第二个循环从i + 1到结束运行(循环计数器j)，第三个循环从j + 1到结束运行(循环计数器k)
  3. 这些循环的计数器表示三元组的3个元素的索引。
  4. 求出ith，jth和kth元素的总和。如果总和等于给定的总和。打印三元组并中断。
  5. 如果没有三联体，则打印不存在三联体。
• 执行：

C++

#include 
using namespace std;
 
// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
bool find3Numbers(int A[], int arr_size, int sum)
{
    int l, r;
 
    // Fix the first element as A[i]
    for (int i = 0; i < arr_size - 2; i++)
    {
 
        // Fix the second element as A[j]
        for (int j = i + 1; j < arr_size - 1; j++)
        {
 
            // Now look for the third number
            for (int k = j + 1; k < arr_size; k++)
            {
                if (A[i] + A[j] + A[k] == sum)
                {
                    cout << "Triplet is " << A[i] <<
                        ", " << A[j] << ", " << A[k];
                    return true;
                }
            }
        }
    }
 
    // If we reach here, then no triplet was found
    return false;
}
 
/* Driver code */
int main()
{
    int A[] = { 1, 4, 45, 6, 10, 8 };
    int sum = 22;
    int arr_size = sizeof(A) / sizeof(A[0]);
    find3Numbers(A, arr_size, sum);
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

#include 
 
// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
bool find3Numbers(int A[], int arr_size, int sum)
{
    int l, r;
 
    // Fix the first element as A[i]
    for (int i = 0; i < arr_size - 2; i++) {
 
        // Fix the second element as A[j]
        for (int j = i + 1; j < arr_size - 1; j++) {
 
            // Now look for the third number
            for (int k = j + 1; k < arr_size; k++) {
                if (A[i] + A[j] + A[k] == sum) {
                    printf("Triplet is %d, %d, %d",
                           A[i], A[j], A[k]);
                    return true;
                }
            }
        }
    }
 
    // If we reach here, then no triplet was found
    return false;
}
 
/* Driver program to test above function */
int main()
{
    int A[] = { 1, 4, 45, 6, 10, 8 };
    int sum = 22;
    int arr_size = sizeof(A) / sizeof(A[0]);
    find3Numbers(A, arr_size, sum);
    return 0;
}

Java

// Java program to find a triplet
class FindTriplet {
 
    // returns true if there is triplet with sum equal
    // to 'sum' present in A[]. Also, prints the triplet
    boolean find3Numbers(int A[], int arr_size, int sum)
    {
        int l, r;
 
        // Fix the first element as A[i]
        for (int i = 0; i < arr_size - 2; i++) {
 
            // Fix the second element as A[j]
            for (int j = i + 1; j < arr_size - 1; j++) {
 
                // Now look for the third number
                for (int k = j + 1; k < arr_size; k++) {
                    if (A[i] + A[j] + A[k] == sum) {
                        System.out.print("Triplet is " + A[i] + ", " + A[j] + ", " + A[k]);
                        return true;
                    }
                }
            }
        }
 
        // If we reach here, then no triplet was found
        return false;
    }
 
    // Driver program to test above functions
    public static void main(String[] args)
    {
        FindTriplet triplet = new FindTriplet();
        int A[] = { 1, 4, 45, 6, 10, 8 };
        int sum = 22;
        int arr_size = A.length;
 
        triplet.find3Numbers(A, arr_size, sum);
    }
}

Python3

# Python3 program to find a triplet
# that sum to a given value
 
# returns true if there is triplet with
# sum equal to 'sum' present in A[].
# Also, prints the triplet
def find3Numbers(A, arr_size, sum):
 
    # Fix the first element as A[i]
    for i in range( 0, arr_size-2):
 
        # Fix the second element as A[j]
        for j in range(i + 1, arr_size-1):
             
            # Now look for the third number
            for k in range(j + 1, arr_size):
                if A[i] + A[j] + A[k] == sum:
                    print("Triplet is", A[i],
                          ", ", A[j], ", ", A[k])
                    return True
     
    # If we reach here, then no
    # triplet was found
    return False
 
# Driver program to test above function
A = [1, 4, 45, 6, 10, 8]
sum = 22
arr_size = len(A)
find3Numbers(A, arr_size, sum)
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# program to find a triplet
// that sum to a given value
using System;
 
class GFG {
    // returns true if there is
    // triplet with sum equal
    // to 'sum' present in A[].
    // Also, prints the triplet
    static bool find3Numbers(int[] A,
                             int arr_size,
                             int sum)
    {
        // Fix the first
        // element as A[i]
        for (int i = 0;
             i < arr_size - 2; i++) {
 
            // Fix the second
            // element as A[j]
            for (int j = i + 1;
                 j < arr_size - 1; j++) {
 
                // Now look for
                // the third number
                for (int k = j + 1;
                     k < arr_size; k++) {
                    if (A[i] + A[j] + A[k] == sum) {
                        Console.WriteLine("Triplet is " + A[i] + ", " + A[j] + ", " + A[k]);
                        return true;
                    }
                }
            }
        }
 
        // If we reach here,
        // then no triplet was found
        return false;
    }
 
    // Driver Code
    static public void Main()
    {
        int[] A = { 1, 4, 45, 6, 10, 8 };
        int sum = 22;
        int arr_size = A.Length;
 
        find3Numbers(A, arr_size, sum);
    }
}
 
// This code is contributed by m_kit

Javascript

 
// Javascript program to find a triplet
// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
function find3Numbers(A, arr_size, sum)
{
    let l, r;
 
    // Fix the first element as A[i]
    for (let i = 0; i < arr_size - 2; i++)
    {
 
        // Fix the second element as A[j]
        for (let j = i + 1; j < arr_size - 1; j++)
        {
 
            // Now look for the third number
            for (let k = j + 1; k < arr_size; k++)
            {
                if (A[i] + A[j] + A[k] == sum)
                {
                    document.write("Triplet is " + A[i] +
                        ", " + A[j] + ", " + A[k]);
                    return true;
                }
            }
        }
    }
 
    // If we reach here, then no triplet was found
    return false;
}
 
/* Driver code */
  
    let A = [ 1, 4, 45, 6, 10, 8 ];
    let sum = 22;
    let arr_size = A.length;
    find3Numbers(A, arr_size, sum);
     
// This code is contributed by Mayank Tyagi
 

输出

Triplet is 4, 10, 8

• 复杂度分析：
  • 时间复杂度： O(n ³ )。
    有三个遍历数组的嵌套循环，因此时间复杂度为O(n ^ 3)
  • 空间复杂度： O(1)。
    由于不需要额外的空间。

方法2：此方法使用排序以提高代码的效率。

• 方法：通过对数组进行排序，可以提高算法的效率。这种有效的方法使用了两指针技术。遍历数组并修复三元组的第一个元素。现在，使用“两指针”算法查找是否存在一对总和等于x – array [i]的对。两个指针算法需要线性时间，因此它比嵌套循环要好。
• 算法 ：
  1. 对给定的数组进行排序。
  2. 遍历数组并修复可能的三元组的第一个元素arr [i]。
  3. 然后固定两个指针，一个指向i + 1，另一个指向n –1。然后看一下总和，
     1. 如果总和小于所需总和，则递增第一个指针。
     2. 否则，如果总和较大，请减小结束指针以减少总和。
     3. 否则，如果两个指针处的元素之和等于给定的和，则打印三元组并中断。
• 执行：

C++

// C++ program to find a triplet
#include 
using namespace std;
 
// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
bool find3Numbers(int A[], int arr_size, int sum)
{
    int l, r;
 
    /* Sort the elements */
    sort(A, A + arr_size);
 
    /* Now fix the first element one by one and find the
       other two elements */
    for (int i = 0; i < arr_size - 2; i++) {
 
        // To find the other two elements, start two index
        // variables from two corners of the array and move
        // them toward each other
        l = i + 1; // index of the first element in the
        // remaining elements
 
        r = arr_size - 1; // index of the last element
        while (l < r) {
            if (A[i] + A[l] + A[r] == sum) {
                printf("Triplet is %d, %d, %d", A[i],
                       A[l], A[r]);
                return true;
            }
            else if (A[i] + A[l] + A[r] < sum)
                l++;
            else // A[i] + A[l] + A[r] > sum
                r--;
        }
    }
 
    // If we reach here, then no triplet was found
    return false;
}
 
/* Driver program to test above function */
int main()
{
    int A[] = { 1, 4, 45, 6, 10, 8 };
    int sum = 22;
    int arr_size = sizeof(A) / sizeof(A[0]);
 
    find3Numbers(A, arr_size, sum);
 
    return 0;
}

Java

// Java program to find a triplet
class FindTriplet {
 
    // returns true if there is triplet with sum equal
    // to 'sum' present in A[]. Also, prints the triplet
    boolean find3Numbers(int A[], int arr_size, int sum)
    {
        int l, r;
 
        /* Sort the elements */
        quickSort(A, 0, arr_size - 1);
 
        /* Now fix the first element one by one and find the
           other two elements */
        for (int i = 0; i < arr_size - 2; i++) {
 
            // To find the other two elements, start two index variables
            // from two corners of the array and move them toward each
            // other
            l = i + 1; // index of the first element in the remaining elements
            r = arr_size - 1; // index of the last element
            while (l < r) {
                if (A[i] + A[l] + A[r] == sum) {
                    System.out.print("Triplet is " + A[i] + ", " + A[l] + ", " + A[r]);
                    return true;
                }
                else if (A[i] + A[l] + A[r] < sum)
                    l++;
 
                else // A[i] + A[l] + A[r] > sum
                    r--;
            }
        }
 
        // If we reach here, then no triplet was found
        return false;
    }
 
    int partition(int A[], int si, int ei)
    {
        int x = A[ei];
        int i = (si - 1);
        int j;
 
        for (j = si; j <= ei - 1; j++) {
            if (A[j] <= x) {
                i++;
                int temp = A[i];
                A[i] = A[j];
                A[j] = temp;
            }
        }
        int temp = A[i + 1];
        A[i + 1] = A[ei];
        A[ei] = temp;
        return (i + 1);
    }
 
    /* Implementation of Quick Sort
    A[] --> Array to be sorted
    si  --> Starting index
    ei  --> Ending index
     */
    void quickSort(int A[], int si, int ei)
    {
        int pi;
 
        /* Partitioning index */
        if (si < ei) {
            pi = partition(A, si, ei);
            quickSort(A, si, pi - 1);
            quickSort(A, pi + 1, ei);
        }
    }
 
    // Driver program to test above functions
    public static void main(String[] args)
    {
        FindTriplet triplet = new FindTriplet();
        int A[] = { 1, 4, 45, 6, 10, 8 };
        int sum = 22;
        int arr_size = A.length;
 
        triplet.find3Numbers(A, arr_size, sum);
    }
}

Python3

# Python3 program to find a triplet
 
# returns true if there is triplet
# with sum equal to 'sum' present
# in A[]. Also, prints the triplet
def find3Numbers(A, arr_size, sum):
 
    # Sort the elements
    A.sort()
 
    # Now fix the first element
    # one by one and find the
    # other two elements
    for i in range(0, arr_size-2):
     
 
        # To find the other two elements,
        # start two index variables from
        # two corners of the array and
        # move them toward each other
         
        # index of the first element
        # in the remaining elements
        l = i + 1
         
        # index of the last element
        r = arr_size-1
        while (l < r):
         
            if( A[i] + A[l] + A[r] == sum):
                print("Triplet is", A[i],
                     ', ', A[l], ', ', A[r]);
                return True
             
            elif (A[i] + A[l] + A[r] < sum):
                l += 1
            else: # A[i] + A[l] + A[r] > sum
                r -= 1
 
    # If we reach here, then
    # no triplet was found
    return False
 
# Driver program to test above function
A = [1, 4, 45, 6, 10, 8]
sum = 22
arr_size = len(A)
 
find3Numbers(A, arr_size, sum)
 
# This is contributed by Smitha Dinesh Semwal

C＃

// C# program to find a triplet
using System;
 
class GFG {
 
    // returns true if there is triplet
    // with sum equal to 'sum' present
    // in A[]. Also, prints the triplet
    bool find3Numbers(int[] A, int arr_size,
                      int sum)
    {
        int l, r;
 
        /* Sort the elements */
        quickSort(A, 0, arr_size - 1);
 
        /* Now fix the first element
    one by one and find the
    other two elements */
        for (int i = 0; i < arr_size - 2; i++) {
 
            // To find the other two elements,
            // start two index variables from
            // two corners of the array and
            // move them toward each other
            l = i + 1; // index of the first element
            // in the remaining elements
            r = arr_size - 1; // index of the last element
            while (l < r) {
                if (A[i] + A[l] + A[r] == sum) {
                    Console.Write("Triplet is " + A[i] + ", " + A[l] + ", " + A[r]);
                    return true;
                }
                else if (A[i] + A[l] + A[r] < sum)
                    l++;
 
                else // A[i] + A[l] + A[r] > sum
                    r--;
            }
        }
 
        // If we reach here, then
        // no triplet was found
        return false;
    }
 
    int partition(int[] A, int si, int ei)
    {
        int x = A[ei];
        int i = (si - 1);
        int j;
 
        for (j = si; j <= ei - 1; j++) {
            if (A[j] <= x) {
                i++;
                int temp = A[i];
                A[i] = A[j];
                A[j] = temp;
            }
        }
        int temp1 = A[i + 1];
        A[i + 1] = A[ei];
        A[ei] = temp1;
        return (i + 1);
    }
 
    /* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
    void quickSort(int[] A, int si, int ei)
    {
        int pi;
 
        /* Partitioning index */
        if (si < ei) {
            pi = partition(A, si, ei);
            quickSort(A, si, pi - 1);
            quickSort(A, pi + 1, ei);
        }
    }
 
    // Driver Code
    static void Main()
    {
        GFG triplet = new GFG();
        int[] A = new int[] { 1, 4, 45, 6, 10, 8 };
        int sum = 22;
        int arr_size = A.Length;
 
        triplet.find3Numbers(A, arr_size, sum);
    }
}
 
// This code is contributed by mits

的PHP

 sum
                $r--;
        }
    }
 
    // If we reach here, then
    // no triplet was found
    return false;
}
 
// Driver Code
$A = array (1, 4, 45, 6, 10, 8);
$sum = 22;
$arr_size = sizeof($A);
 
find3Numbers($A, $arr_size, $sum);
 
// This code is contributed by ajit
?>

输出

Triplet is 4, 8, 10

• 复杂度分析：
  • 时间复杂度： O(N ^ 2)。
    只有两个嵌套循环遍历该数组，因此时间复杂度为O(n ^ 2)。两个指针算法需要O(n)时间，并且可以使用另一个嵌套遍历来固定第一个元素。
  • 空间复杂度： O(1)。
    由于不需要额外的空间。

方法3 ：这是基于哈希的解决方案。

• 方法：这种方法使用了额外的空间，但是比两个指针方法更简单。从头到尾运行两个循环，外循环，从i + 1到结束运行内循环。创建一个哈希图或设置为将元素存储在i + 1到j-1之间。因此，如果给定的总和为x，请检查集合中是否存在等于x – arr [i] – arr [j]的数字。如果是，请打印三元组。
• 算法：
  1. 从头到尾遍历数组。 (循环计数器i)
  2. 创建一个HashMap或设置为存储唯一对。
  3. 从i + 1到数组末尾运行另一个循环。 (循环计数器j)
  4. 如果集合中有一个元素等于x- arr [i] – arr [j]，则打印三元组(arr [i]，arr [j]，x-arr [i] -arr [j] )并打破
  5. 在集合中插入第j个元素。
• 执行：

C++

// C++ program to find a triplet using Hashing
#include 
using namespace std;
 
// returns true if there is triplet with sum equal
// to 'sum' present in A[]. Also, prints the triplet
bool find3Numbers(int A[], int arr_size, int sum)
{
    // Fix the first element as A[i]
    for (int i = 0; i < arr_size - 2; i++)
    {
 
        // Find pair in subarray A[i+1..n-1]
        // with sum equal to sum - A[i]
        unordered_set s;
        int curr_sum = sum - A[i];
        for (int j = i + 1; j < arr_size; j++)
        {
            if (s.find(curr_sum - A[j]) != s.end())
            {
                printf("Triplet is %d, %d, %d", A[i],
                       A[j], curr_sum - A[j]);
                return true;
            }
            s.insert(A[j]);
        }
    }
 
    // If we reach here, then no triplet was found
    return false;
}
 
/* Driver program to test above function */
int main()
{
    int A[] = { 1, 4, 45, 6, 10, 8 };
    int sum = 22;
    int arr_size = sizeof(A) / sizeof(A[0]);
 
    find3Numbers(A, arr_size, sum);
 
    return 0;
}

Java

// Java program to find a triplet using Hashing
import java.util.*;
 
class GFG {
 
    // returns true if there is triplet
    // with sum equal to 'sum' present
    // in A[]. Also, prints the triplet
    static boolean find3Numbers(int A[],
                                int arr_size, int sum)
    {
        // Fix the first element as A[i]
        for (int i = 0; i < arr_size - 2; i++) {
 
            // Find pair in subarray A[i+1..n-1]
            // with sum equal to sum - A[i]
            HashSet s = new HashSet();
            int curr_sum = sum - A[i];
            for (int j = i + 1; j < arr_size; j++)
            {
                if (s.contains(curr_sum - A[j]))
                {
                    System.out.printf("Triplet is %d,
                                        %d, %d", A[i],
                                      A[j], curr_sum - A[j]);
                    return true;
                }
                s.add(A[j]);
            }
        }
 
        // If we reach here, then no triplet was found
        return false;
    }
 
    /* Driver code */
    public static void main(String[] args)
    {
        int A[] = { 1, 4, 45, 6, 10, 8 };
        int sum = 22;
        int arr_size = A.length;
 
        find3Numbers(A, arr_size, sum);
    }
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 program to find a triplet using Hashing
# returns true if there is triplet with sum equal
# to 'sum' present in A[]. Also, prints the triplet
def find3Numbers(A, arr_size, sum):
    for i in range(0, arr_size-1):
        # Find pair in subarray A[i + 1..n-1]
        # with sum equal to sum - A[i]
        s = set()
        curr_sum = sum - A[i]
        for j in range(i + 1, arr_size):
            if (curr_sum - A[j]) in s:
                print("Triplet is", A[i],
                        ", ", A[j], ", ", curr_sum-A[j])
                return True
            s.add(A[j])
     
    return False
 
# Driver program to test above function
A = [1, 4, 45, 6, 10, 8]
sum = 22
arr_size = len(A)
find3Numbers(A, arr_size, sum)
 
# This is contributed by Yatin gupta

C＃

// C# program to find a triplet using Hashing
using System;
using System.Collections.Generic;
public class GFG {
 
    // returns true if there is triplet
    // with sum equal to 'sum' present
    // in A[]. Also, prints the triplet
    static bool find3Numbers(int[] A,
                             int arr_size, int sum)
    {
        // Fix the first element as A[i]
        for (int i = 0; i < arr_size - 2; i++) {
 
            // Find pair in subarray A[i+1..n-1]
            // with sum equal to sum - A[i]
            HashSet s = new HashSet();
            int curr_sum = sum - A[i];
            for (int j = i + 1; j < arr_size; j++)
            {
                if (s.Contains(curr_sum - A[j]))
                {
                    Console.Write("Triplet is {0}, {1}, {2}", A[i],
                                  A[j], curr_sum - A[j]);
                    return true;
                }
                s.Add(A[j]);
            }
        }
 
        // If we reach here, then no triplet was found
        return false;
    }
 
    /* Driver code */
    public static void Main()
    {
        int[] A = { 1, 4, 45, 6, 10, 8 };
        int sum = 22;
        int arr_size = A.Length;
 
        find3Numbers(A, arr_size, sum);
    }
}
 
/* This code contributed by PrinciRaj1992 */

输出

Triplet is 4, 8, 10

• 复杂度分析：
  • 时间复杂度： O(N ^ 2)。
    只有两个嵌套循环遍历该数组，因此时间复杂度为O(n ^ 2)。
  • 空间复杂度： O(N)。
    由于不需要额外的空间。