求和数组之谜

给定n个整数的数组arr []，构造一个总和数组sum [](大小相同)，以使sum [i]等于arr []中除arr [i]以外的所有元素的和。在没有减法运算符并在O(n)中求解。

例子：

Input : arr[] = {3, 6, 4, 8, 9}
Output : sum[] = {27, 24, 26, 22, 21}

Input : arr[] = {4, 5, 7, 3, 10, 1}
Output : sum[] = {26, 25, 23, 27, 20, 29}

算法：
1)构造一个临时数组leftSum []，使leftSum [i]包含arr [i]左侧所有元素的总和，但不包括arr [i]。
2)构造另一个临时数组rightSum []，以使rightSum [i]包含arr [i]右边的所有元素之和，不包括arr [i]。
3)要获得sum []，请将left []和right []相加。

C/C++

// C++ implementation of above approach
#include 
using namespace std;
  
void sumArray(int arr[], int n)
{
    /* Allocate memory for temporary arrays leftSum[],
       rightSum[] and Sum[]*/
    int leftSum[n], rightSum[n], Sum[n], i, j;
  
    /* Left most element of left array is always 0 */
    leftSum[0] = 0;
  
    /* Rightmost most element of right 
       array is always 0 */
    rightSum[n - 1] = 0;
  
    /* Construct the left array*/
    for (i = 1; i < n; i++)
        leftSum[i] = arr[i - 1] + leftSum[i - 1];
  
    /* Construct the right array*/
    for (j = n - 2; j >= 0; j--)
        rightSum[j] = arr[j + 1] + rightSum[j + 1];
  
    /* Construct the sum array using 
        left[] and right[] */
    for (i = 0; i < n; i++)
        Sum[i] = leftSum[i] + rightSum[i];
  
    /* print the constructed prod array */
    for (i = 0; i < n; i++)
        cout << Sum[i] << " ";
}
  
/* Driver program to test above functions */
int main()
{
    int arr[] = { 3, 6, 4, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    sumArray(arr, n);
  
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
  
class Geeks {
    public static void sumArray(int arr[], int n)
    {
        /* Allocate memory for temporary arrays 
           leftSum[], rightSum[] and Sum[]*/
        int leftSum[] = new int[n];
        int rightSum[] = new int[n];
        int Sum[] = new int[n];
  
        int i = 0, j = 0;
  
        /* Left most element of left array is 
           always 0 */
        leftSum[0] = 0;
  
        /* Right most element of rght array 
           is always 0 */
        rightSum[n - 1] = 0;
  
        /* Construct the left array*/
        for (i = 1; i < n; i++)
            leftSum[i] = arr[i - 1] + leftSum[i - 1];
  
        /* Construct the right array*/
        for (j = n - 2; j >= 0; j--)
            rightSum[j] = arr[j + 1] + rightSum[j + 1];
  
        /* Construct the sum array using 
        left[] and right[] */
        for (i = 0; i < n; i++)
            Sum[i] = leftSum[i] + rightSum[i];
  
        /*print the sum array*/
        for (i = 0; i < n; i++)
            System.out.print(Sum[i] + " ");
    }
  
    /* Driver function to test above function*/
    public static void main(String[] args)
    {
        int arr[] = { 3, 6, 4, 8, 9 };
        int n = arr.length;
        sumArray(arr, n);
    }
}

Python3

# Python3 implementation of above approach
  
def sumArray(arr, n):
  
    # Allocate memory for temporary arrays 
    # leftSum[], rightSum[] and Sum[]
    leftSum = [0 for i in range(n)]
    rightSum = [0 for i in range(n)]
    Sum = [0 for i in range(n)]
    i, j = 0, 0
  
    # Left most element of left 
    # array is always 0 
    leftSum[0] = 0
  
    # Rightmost most element of right 
    # array is always 0 
    rightSum[n - 1] = 0
  
    # Construct the left array
    for i in range(1, n):
        leftSum[i] = arr[i - 1] + leftSum[i - 1]
  
    # Construct the right array
    for j in range(n - 2, -1, -1):
        rightSum[j] = arr[j + 1] + rightSum[j + 1]
  
    # Construct the sum array using 
    # left[] and right[] 
    for i in range(0, n):
        Sum[i] = leftSum[i] + rightSum[i]
  
    # print the constructed prod array 
    for i in range(n):
        print(Sum[i], end = " ")
  
  
# Driver Code
arr = [3, 6, 4, 8, 9] 
n = len(arr)
sumArray(arr, n)
  
# This code is contributed by 
# mohit kumar 29

C#

// C# implementation of above approach
  
using System ;
  
class Geeks { 
    public static void sumArray(int []arr, int n) 
    { 
        /* Allocate memory for temporary arrays 
        leftSum[], rightSum[] and Sum[]*/
        int []leftSum = new int[n]; 
        int []rightSum = new int[n]; 
        int []Sum = new int[n]; 
  
        int i = 0, j = 0; 
  
        /* Left most element of left array is 
        always 0 */
        leftSum[0] = 0; 
  
        /* Right most element of rght array 
        is always 0 */
        rightSum[n - 1] = 0; 
  
        /* Construct the left array*/
        for (i = 1; i < n; i++) 
            leftSum[i] = arr[i - 1] + leftSum[i - 1]; 
  
        /* Construct the right array*/
        for (j = n - 2; j >= 0; j--) 
            rightSum[j] = arr[j + 1] + rightSum[j + 1]; 
  
        /* Construct the sum array using 
        left[] and right[] */
        for (i = 0; i < n; i++) 
            Sum[i] = leftSum[i] + rightSum[i]; 
  
        /*print the sum array*/
        for (i = 0; i < n; i++) 
            Console.Write(Sum[i] + " "); 
    } 
  
    /* Driver function to test above function*/
    public static void Main() 
    { 
        int []arr = { 3, 6, 4, 8, 9 }; 
        int n = arr.Length; 
        sumArray(arr, n); 
    } 
    // This code is contributed by Ryuga
}

PHP

= 0; $j--)
        $rightSum[$j] = $arr[$j + 1] +
                        $rightSum[$j + 1];
  
    /* Construct the sum array using 
        left[] and right[] */
    for ($i = 0; $i < $n; $i++)
        $Sum[$i] = $leftSum[$i] + $rightSum[$i];
  
    /* print the constructed prod array */
    for ($i = 0; $i < $n; $i++)
        echo $Sum[$i]." ";
}
  
// Driver Code
$arr = array( 3, 6, 4, 8, 9 );
$n = count($arr);
sumArray($arr, $n);
  
// This code is contributed
// by chandan_jnu
?>

输出：

27 24 26 22 21

时间复杂度： O(n)
空间复杂度： O(n)
辅助空间： O(n)

可以优化上述方法以在O(1)辅助空间中工作。

C / C++

// C++ implementation of above approach
#include 
using namespace std;
  
void sumArray(int arr[], int n)
{
    int i, temp = 0;
  
    /* Allocate memory for the sum array */
    int Sum[n];
  
    /* Initialize the sum array as 0 */
    memset(Sum, 0, n);
  
    /* In this loop, temp variable contains 
       sum of elements on left side excluding
       arr[i] */
    for (i = 0; i < n; i++) {
        Sum[i] = temp;
        temp += arr[i];
    }
  
    /* Initialize temp to 0 for sum on right
       side */
    temp = 0;
  
    /* In this loop, temp variable contains
       sum of  elements on right side excluding
        arr[i] */
    for (i = n - 1; i >= 0; i--) {
        Sum[i] += temp;
        temp += arr[i];
    }
  
    for (i = 0; i < n; i++)
        cout << Sum[i] << " ";
}
  
/* Driver program to test above function */
int main()
{
    int arr[] = { 3, 6, 4, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    sumArray(arr, n);
  
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
  
class Geeks {
    public static void sumArray(int arr[], int n)
    {
        int i = 0, temp = 0;
        int Sum[] = new int[n];
  
        Arrays.fill(Sum, 0);
  
        /* In this loop, temp variable contains
           sum of elements on left side excluding
           arr[i] */
        for (i = 0; i < n; i++) {
            Sum[i] = temp;
            temp += arr[i];
        }
  
        /* Initialize temp to 0 for sum on right side */
        temp = 0;
  
        /* In this loop, temp variable contains
           sum of elements on right side excluding
           arr[i] */
        for (i = n - 1; i >= 0; i--) {
            Sum[i] += temp;
            temp += arr[i];
        }
  
        for (i = 0; i < n; i++)
            System.out.print(Sum[i] + " ");
    }
  
    /* Driver function to test above function*/
    public static void main(String[] args)
    {
        int arr[] = { 3, 6, 4, 8, 9 };
        int n = arr.length;
        sumArray(arr, n);
    }
}

Python3

# Python3 implementation of above approach
def sumArray(arr, n):
  
    i, temp = 0, 0
  
    # Allocate memory for the sum array */
    Sum = [0 for i in range(n)]
  
    '''In this loop, temp variable contains 
    sum of elements on left side excluding
    arr[i] '''
    for i in range(n):
        Sum[i] = temp
        temp += arr[i]
      
    # Initialize temp to 0 for sum 
    # on right side */
    temp = 0
  
    ''' In this loop, temp variable contains
        sum of elements on right side excluding
        arr[i] */'''
    for i in range(n - 1, -1, -1):
        Sum[i] += temp
        temp += arr[i]
      
    for i in range(n):
        print(Sum[i], end = " ")
  
# Driver Code
arr = [ 3, 6, 4, 8, 9 ]
n = len(arr)
sumArray(arr, n)
  
# This code is contributed by
# Mohit Kumar 29

C＃

// C# implementation of above approach
using System;
class Geeks {
    public static void sumArray(int []arr, int n)
    {
        int i = 0, temp = 0;
        int []Sum = new int[n];
        for( i=0;i= 0; i--) {
            Sum[i] += temp;
            temp += arr[i];
        }
  
        for (i = 0; i < n; i++)
            Console.Write(Sum[i] + " ");
    }
  
    /* Driver function to test above function*/
    public static void Main()
    {
        int []arr = { 3, 6, 4, 8, 9 };
        int n = arr.Length;
        sumArray(arr, n);
    }
}
// This code is contributed by inder_verma..

的PHP

= 0; $i--) 
    {
        $Sum[$i] += $temp;
        $temp += $arr[$i];
    }
  
    for ($i = 0; $i < $n; $i++)
        echo $Sum[$i] . " ";
}
  
// Driver Code
$arr = array( 3, 6, 4, 8, 9 );
$n = count($arr);
sumArray($arr, $n);
  
// This code is contributed by
// chandan_jnu
?>

输出：

27 24 26 22 21

时间复杂度： O(n)
辅助空间： O(1)