找到四个总和为给定值的元素|设置1(n ^ 3个解决方案)

给定一个整数数组，找到该数组中四个和的总和等于给定值X的所有组合。
例如，如果给定的数组是{10，2，3，4，5，9，7，8}且X = 23，则您的函数应打印“ 3 5 7 8”(3 + 5 + 7 + 8 = 23)。

一个朴素的解决方案是生成所有可能的四元组，并将每个四元组的和与X进行比较。以下代码使用四个嵌套循环实现此简单方法

C++

// C++ program for naive solution to
// print all combination of 4 elements
// in A[] with sum equal to X
#include 
using namespace std;
 
/* A naive solution to print all combination
of 4 elements in A[]with sum equal to X */
void findFourElements(int A[], int n, int X)
{
     
// Fix the first element and find other three
for (int i = 0; i < n - 3; i++)
{
    // Fix the second element and find other two
    for (int j = i + 1; j < n - 2; j++)
    {
         
        // Fix the third element and find the fourth
        for (int k = j + 1; k < n - 1; k++)
        {
            // find the fourth
            for (int l = k + 1; l < n; l++)
            if (A[i] + A[j] + A[k] + A[l] == X)
                cout << A[i] <<", " << A[j]
                     << ", " << A[k] << ", " << A[l];
        }
    }
}
}
 
// Driver Code
int main()
{
    int A[] = {10, 20, 30, 40, 1, 2};
    int n = sizeof(A) / sizeof(A[0]);
    int X = 91;
    findFourElements (A, n, X);
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

#include 
 
/* A naive solution to print all combination of 4 elements in A[]
  with sum equal to X */
void findFourElements(int A[], int n, int X)
{
  // Fix the first element and find other three
  for (int i = 0; i < n-3; i++)
  {
    // Fix the second element and find other two
    for (int j = i+1; j < n-2; j++)
    {
      // Fix the third element and find the fourth
      for (int k = j+1; k < n-1; k++)
      {
        // find the fourth
        for (int l = k+1; l < n; l++)
           if (A[i] + A[j] + A[k] + A[l] == X)
              printf("%d, %d, %d, %d", A[i], A[j], A[k], A[l]);
      }
    }
  }
}
 
// Driver program to test above function
int main()
{
    int A[] = {10, 20, 30, 40, 1, 2};
    int n = sizeof(A) / sizeof(A[0]);
    int X = 91;
    findFourElements (A, n, X);
    return 0;
}

Java

class FindFourElements
{
 
    /* A naive solution to print all combination of 4 elements in A[]
       with sum equal to X */
    void findFourElements(int A[], int n, int X)
    {
        // Fix the first element and find other three
        for (int i = 0; i < n - 3; i++)
        {
            // Fix the second element and find other two
            for (int j = i + 1; j < n - 2; j++)
            {
                // Fix the third element and find the fourth
                for (int k = j + 1; k < n - 1; k++)
                {
                    // find the fourth
                    for (int l = k + 1; l < n; l++)
                    {
                        if (A[i] + A[j] + A[k] + A[l] == X)
                            System.out.print(A[i]+" "+A[j]+" "+A[k]
                                                                 +" "+A[l]);
                    }
                }
            }
        }
    }
 
    // Driver program to test above functions
    public static void main(String[] args)
    {
        FindFourElements findfour = new FindFourElements();
        int A[] = {10, 20, 30, 40, 1, 2};
        int n = A.length;
        int X = 91;
        findfour.findFourElements(A, n, X);
    }
}

Python3

# A naive solution to print all combination
# of 4 elements in A[] with sum equal to X
def findFourElements(A, n, X):
     
    # Fix the first element and find
    # other three
    for i in range(0,n-3):
         
        # Fix the second element and
        # find other two
        for j in range(i+1,n-2):
             
            # Fix the third element
            # and find the fourth
            for k in range(j+1,n-1):
                 
                # find the fourth
                for l in range(k+1,n):
                     
                    if A[i] + A[j] + A[k] + A[l] == X:
                        print ("%d, %d, %d, %d"
                        %( A[i], A[j], A[k], A[l]))
 
# Driver program to test above function
A = [10, 2, 3, 4, 5, 9, 7, 8]
n = len(A)
X = 23
findFourElements (A, n, X)
 
# This code is contributed by shreyanshi_arun

C#

// C# program for naive solution to
// print all combination of 4 elements
// in A[] with sum equal to X
using System;
 
class FindFourElements
{
    void findFourElements(int []A, int n, int X)
    {
        // Fix the first element and find other three
        for (int i = 0; i < n - 3; i++)
        {
            // Fix the second element and find other two
            for (int j = i + 1; j < n - 2; j++)
            {
                // Fix the third element and find the fourth
                for (int k = j + 1; k < n - 1; k++)
                {
                    // find the fourth
                    for (int l = k + 1; l < n; l++)
                    {
                        if (A[i] + A[j] + A[k] + A[l] == X)
                        Console.Write(A[i] + " " + A[j] +
                                " " + A[k] + " " + A[l]);
                    }
                }
            }
        }
    }
 
    // Driver program to test above functions
    public static void Main()
    {
        FindFourElements findfour = new FindFourElements();
        int []A = {10, 20, 30, 40, 1, 2};
        int n = A.Length;
        int X = 91;
        findfour.findFourElements(A, n, X);
    }
}
 
// This code is contributed by nitin mittal

Javascript

 
// Javascript program for the above approach
 
/* A naive solution to print all combination
of 4 elements in A[]with sum equal to X */
function findFourElements(A, n, X)
{
     
// Fix the first element and find other three
for (let i = 0; i < n - 3; i++)
{
    // Fix the second element and find other two
    for (let j = i + 1; j < n - 2; j++)
    {
         
        // Fix the third element and find the fourth
        for (let k = j + 1; k < n - 1; k++)
        {
            // find the fourth
            for (let l = k + 1; l < n; l++)
            if (A[i] + A[j] + A[k] + A[l] == X)
                document.write(A[i]+", "+A[j]+", "+A[k] +", "+A[l]);
        }
    }
}
}
 
// Driver Code
 
    let A = [10, 20, 30, 40, 1, 2];
    let n = A.length;
    let X = 91;
    findFourElements (A, n, X);
 

输出：

20, 30, 40, 1

时间复杂度：O(n ^ 4)
的时间复杂度可以通过使用排序作为预处理步骤，并且然后使用此信息的方法1，以减少循环的改进，以O(N ^ 3)。
以下是详细步骤。
1)对输入数组进行排序。
2)将第一个元素固定为A [i]，其中i从0到n–3。固定四元组的第一个元素后，将第二个元素固定为A [j]，其中j从i + 1到n-2变化。使用本文的方法1查找O(n)时间中剩余的两个元素
以下是O(n ^ 3)解决方案的实现。

C++

// C++ program for to  print all combination
// of 4 elements in A[] with sum equal to X
#include
using namespace std;
 
/* Following function is needed
for library function qsort(). */
int compare (const void *a, const void * b)
{
    return ( *(int *)a - *(int *)b );
}
 
/* A sorting based solution to print
all combination of 4 elements in A[]
with sum equal to X */
void find4Numbers(int A[], int n, int X)
{
    int l, r;
 
    // Sort the array in increasing
    // order, using library function
    // for quick sort
    qsort (A, n, sizeof(A[0]), compare);
 
    /* Now fix the first 2 elements
    one by one and find
    the other two elements */
    for (int i = 0; i < n - 3; i++)
    {
        for (int j = i+1; j < n - 2; j++)
        {
            // Initialize two variables as
            // indexes of the first and last
            // elements in the remaining elements
            l = j + 1;
            r = n-1;
 
            // To find the remaining two
            // elements, move the index
            // variables (l & r) toward each other.
            while (l < r)
            {
                if( A[i] + A[j] + A[l] + A[r] == X)
                {
                    cout << A[i]<<", " << A[j] <<
                        ", " << A[l] << ", " << A[r];
                    l++; r--;
                }
                else if (A[i] + A[j] + A[l] + A[r] < X)
                    l++;
                else // A[i] + A[j] + A[l] + A[r] > X
                    r--;
            } // end of while
        } // end of inner for loop
    } // end of outer for loop
}
 
/* Driver code */
int main()
{
    int A[] = {1, 4, 45, 6, 10, 12};
    int X = 21;
    int n = sizeof(A) / sizeof(A[0]);
    find4Numbers(A, n, X);
    return 0;
}
 
// This code is contributed by rathbhupendra

C

# include 
# include 
 
/* Following function is needed for library function qsort(). Refer
   http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
int compare (const void *a, const void * b)
{  return ( *(int *)a - *(int *)b ); }
 
/* A sorting based solution to print all combination of 4 elements in A[]
   with sum equal to X */
void find4Numbers(int A[], int n, int X)
{
    int l, r;
 
    // Sort the array in increasing order, using library
    // function for quick sort
    qsort (A, n, sizeof(A[0]), compare);
 
    /* Now fix the first 2 elements one by one and find
       the other two elements */
    for (int i = 0; i < n - 3; i++)
    {
        for (int j = i+1; j < n - 2; j++)
        {
            // Initialize two variables as indexes of the first and last
            // elements in the remaining elements
            l = j + 1;
            r = n-1;
 
            // To find the remaining two elements, move the index
            // variables (l & r) toward each other.
            while (l < r)
            {
                if( A[i] + A[j] + A[l] + A[r] == X)
                {
                   printf("%d, %d, %d, %d", A[i], A[j],
                                           A[l], A[r]);
                   l++; r--;
                }
                else if (A[i] + A[j] + A[l] + A[r] < X)
                    l++;
                else // A[i] + A[j] + A[l] + A[r] > X
                    r--;
            } // end of while
        } // end of inner for loop
    } // end of outer for loop
}
 
/* Driver program to test above function */
int main()
{
    int A[] = {1, 4, 45, 6, 10, 12};
    int X = 21;
    int n = sizeof(A)/sizeof(A[0]);
    find4Numbers(A, n, X);
    return 0;
}

Java

import java.util.Arrays;
 
 
class FindFourElements
{
    /* A sorting based solution to print all combination of 4 elements in A[]
       with sum equal to X */
    void find4Numbers(int A[], int n, int X)
    {
        int l, r;
 
        // Sort the array in increasing order, using library
        // function for quick sort
        Arrays.sort(A);
 
        /* Now fix the first 2 elements one by one and find
           the other two elements */
        for (int i = 0; i < n - 3; i++)
        {
            for (int j = i + 1; j < n - 2; j++)
            {
                // Initialize two variables as indexes of the first and last
                // elements in the remaining elements
                l = j + 1;
                r = n - 1;
 
                // To find the remaining two elements, move the index
                // variables (l & r) toward each other.
                while (l < r)
                {
                    if (A[i] + A[j] + A[l] + A[r] == X)
                    {
                        System.out.println(A[i]+" "+A[j]+" "+A[l]+" "+A[r]);
                        l++;
                        r--;
                    }
                    else if (A[i] + A[j] + A[l] + A[r] < X)
                        l++;
                    else // A[i] + A[j] + A[l] + A[r] > X
                        r--;
                } // end of while
            } // end of inner for loop
        } // end of outer for loop
    }
 
    // Driver program to test above functions
    public static void main(String[] args)
    {
        FindFourElements findfour = new FindFourElements();
        int A[] = {1, 4, 45, 6, 10, 12};
        int n = A.length;
        int X = 21;
        findfour.find4Numbers(A, n, X);
    }
}
 
// This code has been contributed by Mayank Jaiswal

的Python 3

# A sorting based solution to print all combination
# of 4 elements in A[] with sum equal to X
def find4Numbers(A, n, X):
 
    # Sort the array in increasing order,
    # using library function for quick sort
    A.sort()
 
    # Now fix the first 2 elements one by
    # one and find the other two elements
    for i in range(n - 3):
        for j in range(i + 1, n - 2):
             
            # Initialize two variables as indexes
            # of the first and last elements in
            # the remaining elements
            l = j + 1
            r = n - 1
 
            # To find the remaining two elements,
            # move the index variables (l & r)
            # toward each other.
            while (l < r):
                if(A[i] + A[j] + A[l] + A[r] == X):
                    print(A[i], ",", A[j], ",",
                          A[l], ",", A[r])
                    l += 1
                    r -= 1
                 
                elif (A[i] + A[j] + A[l] + A[r] < X):
                    l += 1
                else: # A[i] + A[j] + A[l] + A[r] > X
                    r -= 1
 
# Driver Code
if __name__ == "__main__":
     
    A = [1, 4, 45, 6, 10, 12]
    X = 21
    n = len(A)
    find4Numbers(A, n, X)
 
# This code is contributed by ita_c

C＃

// C# program for to  print all combination
// of 4 elements in A[] with sum equal to X
using System;
 
class FindFourElements
{
    // A sorting based solution to print all
    // combination of 4 elements in A[] with
    // sum equal to X
    void find4Numbers(int []A, int n, int X)
    {
        int l, r;
 
        // Sort the array in increasing order,
        // using library function for quick sort
        Array.Sort(A);
 
        /* Now fix the first 2 elements one by one
           and find the other two elements */
        for (int i = 0; i < n - 3; i++)
        {
            for (int j = i + 1; j < n - 2; j++)
            {
                // Initialize two variables as indexes of
                // the first and last elements in the
                // remaining elements
                l = j + 1;
                r = n - 1;
 
                // To find the remaining two elements, move the
                // index variables (l & r) toward each other.
                while (l < r)
                {
                    if (A[i] + A[j] + A[l] + A[r] == X)
                    {
                        Console.Write(A[i] + " " + A[j] +
                                " " + A[l] + " " + A[r]);
                        l++;
                        r--;
                    }
                    else if (A[i] + A[j] + A[l] + A[r] < X)
                        l++;
                         
                    else // A[i] + A[j] + A[l] + A[r] > X
                        r--;
                         
                } // end of while
                 
            } // end of inner for loop
             
        } // end of outer for loop
    }
 
    // Driver program to test above functions
    public static void Main()
    {
        FindFourElements findfour = new FindFourElements();
        int []A = {1, 4, 45, 6, 10, 12};
        int n = A.Length;
        int X = 21;
        findfour.find4Numbers(A, n, X);
    }
}
 
// This code has been contributed by nitin mittal

的PHP

 X
                else
                    $r--;
                     
            }
        }
    }
}
 
// Driver Code
$A = array(1, 4, 45, 6, 10, 12);
$n = count($A);
$X = 21;
find4Numbers($A, $n, $X);
 
// This code is contributed
// by nitin mittal
?>

Java脚本

     
    /* A sorting based solution to print all combination of 4 elements in A[]
       with sum equal to X */
    function find4Numbers(A,n,X)
    {
         
        let l, r;
        // Sort the array in increasing order, using library
        // function for quick sort
        A.sort(function(a, b){return a - b});
         
        /* Now fix the first 2 elements one by one and find
           the other two elements */
        for (let i = 0; i < n - 3; i++)
        {
            for (let j = i + 1; j < n - 2; j++)
            {
                // Initialize two variables as indexes of the first and last
                // elements in the remaining elements
                l = j + 1;
                r = n - 1;
  
                // To find the remaining two elements, move the index
                // variables (l & r) toward each other.
                while (l < r)
                {
                    if (A[i] + A[j] + A[l] + A[r] == X)
                    {
                        document.write(A[i]+" "+A[j]+" "+A[l]+" "+A[r]+"<br>");
                        l++;
                        r--;
                    }
                    else if ((A[i] + A[j] + A[l] + A[r]) < X)
                    {   
                        l++;
                    }
                    else // A[i] + A[j] + A[l] + A[r] > X
                    {
                        r--;
                    }
                } // end of while
            }    // end of inner for loop
        }// end of outer for loop
    }
     
    // Driver program to test above functions
    let A= [1, 4, 45, 6, 10, 12];
    let n = A.length;
    let X = 21;
     
    find4Numbers(A, n, X)
     
    // This code is contributed by rag2127
     

输出 ：

1, 4, 6, 10

时间复杂度： O(n ^ 3)