查找给定数组的所有唯一子数组总和。

给定n个正元素数组。子阵列总和定义为特定子阵列所有元素的总和，任务是查找所有唯一子阵列总和的总和。
注意：唯一子数组总和表示没有其他子数组具有相同的总和值。
例子：

Input : arr[] = {3, 4, 5} 
Output : 40 
Explanation: All possible unique sub-array with their sum are as: 
(3), (4), (5), (3+4), (4+5), (3+4+5). Here all are unique so required sum = 40
Input : arr[] = {2, 4, 2} 
Output : 12 
Explanation: All possible unique sub-array with their sum are as: 
(2), (4), (2), (2+4), (4+2), (2+4+2). Here only (4) and (2+4+2) are unique.

方法1(基于排序)
1-计算数组的累加和。
2-将所有子数组总和存储在向量中。
3-对向量排序。
4-将所有重复的子数组总和标记为零
5-计算并返回totalSum。

C++

// C++ for finding sum of all unique
// subarray sum
#include 
using namespace std;
 
// function for finding grandSum
long long int findSubarraySum(int arr[], int n)
{
    int i, j;
 
    // calculate cumulative sum of array
    // cArray[0] will store sum of zero elements
    long long int cArray[n + 1] = { 0 };
    for (i = 0; i < n; i++)
        cArray[i + 1] = cArray[i] + arr[i];
 
    vector subArrSum;
 
    // store all subarray sum in vector
    for (i = 1; i <= n; i++)
        for (j = i; j <= n; j++)
            subArrSum.push_back(cArray[j] -
                                cArray[i - 1]);
 
    // sort the vector
    sort(subArrSum.begin(), subArrSum.end());
 
    // mark all duplicate sub-array
    // sum to zero
    long long totalSum = 0;
    for (i = 0; i < subArrSum.size() - 1; i++)
    {
        if (subArrSum[i] == subArrSum[i + 1])
        {
            j = i + 1;
            while (subArrSum[j] == subArrSum[i] &&
                                   j < subArrSum.size())
            {
                subArrSum[j] = 0;
                j++;
            }
            subArrSum[i] = 0;
        }
    }
 
    // calculate total sum
    for (i = 0; i < subArrSum.size(); i++)
        totalSum += subArrSum[i];
 
    // return totalSum
    return totalSum;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSubarraySum(arr, n);
    return 0;
}

Java

// Java for finding sum of all unique
// subarray sum
import java.util.*;
 
class GFG{
 
// Function for finding grandSum
static int findSubarraySum(int arr[], int n)
{
    int i, j;
 
    // Calculate cumulative sum of array
    // cArray[0] will store sum of zero elements
    int cArray[] = new int[n + 1];
    for(i = 0; i < n; i++)
        cArray[i + 1] = cArray[i] + arr[i];
 
    Vector subArrSum = new Vector();
 
    // Store all subarray sum in vector
    for(i = 1; i <= n; i++)
        for(j = i; j <= n; j++)
            subArrSum.add(cArray[j] -
                          cArray[i - 1]);
 
    // Sort the vector
    Collections.sort(subArrSum);
 
    // Mark all duplicate sub-array
    // sum to zero
    int totalSum = 0;
    for(i = 0; i < subArrSum.size() - 1; i++)
    {
        if (subArrSum.get(i) ==
            subArrSum.get(i + 1))
        {
            j = i + 1;
            while (subArrSum.get(j) ==
                   subArrSum.get(i) &&
               j < subArrSum.size())
            {
                subArrSum.set(j, 0);
                j++;
            }
            subArrSum.set(i, 0);
        }
    }
 
    // Calculate total sum
    for(i = 0; i < subArrSum.size(); i++)
        totalSum += subArrSum.get(i);
 
    // Return totalSum
    return totalSum;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = arr.length;
     
    System.out.print(findSubarraySum(arr, n));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 for finding sum of all
# unique subarray sum
 
# function for finding grandSum
def findSubarraySum(arr, n):
     
    # calculate cumulative sum of array
    # cArray[0] will store sum of zero elements
    cArray = [0 for i in range(n + 1)]
    for i in range(0, n, 1):
        cArray[i + 1] = cArray[i] + arr[i]
 
    subArrSum = []
 
    # store all subarray sum in vector
    for i in range(1, n + 1, 1):
        for j in range(i, n + 1, 1):
            subArrSum.append(cArray[j] -
                             cArray[i - 1])
 
    # sort the vector
    subArrSum.sort(reverse = False)
 
    # mark all duplicate sub-array
    # sum to zero
    totalSum = 0
    for i in range(0, len(subArrSum) - 1, 1):
        if (subArrSum[i] == subArrSum[i + 1]):
            j = i + 1
            while (subArrSum[j] == subArrSum[i] and
                           j < len(subArrSum)):
                subArrSum[j] = 0
                j += 1
            subArrSum[i] = 0
 
    # calculate total sum
    for i in range(0, len(subArrSum), 1):
        totalSum += subArrSum[i]
 
    # return totalSum
    return totalSum
 
# Drivers code
if __name__ == '__main__':
    arr = [3, 2, 3, 1, 4]
    n = len(arr)
    print(findSubarraySum(arr, n))
     
# This code is contributed by
# Sahil_Shelangia

C#

// C# for finding sum of all unique
// subarray sum
using System;
using System.Collections.Generic;
class GFG{
 
// Function for finding grandSum
static int findSubarraySum(int []arr,
                           int n)
{
  int i, j;
 
  // Calculate cumulative sum
  // of array cArray[0] will
  // store sum of zero elements
  int []cArray = new int[n + 1];
   
  for(i = 0; i < n; i++)
    cArray[i + 1] = cArray[i] + arr[i];
 
  List subArrSum = new List();
 
  // Store all subarray sum in vector
  for(i = 1; i <= n; i++)
    for(j = i; j <= n; j++)
      subArrSum.Add(cArray[j] -
                    cArray[i - 1]);
 
  // Sort the vector
  subArrSum.Sort();
 
  // Mark all duplicate
  // sub-array sum to zero
  int totalSum = 0;
  for(i = 0; i < subArrSum.Count - 1; i++)
  {
    if (subArrSum[i] ==
        subArrSum[i + 1])
    {
      j = i + 1;
      while (subArrSum[j] ==
             subArrSum[i] &&
             j < subArrSum.Count)
      {
        subArrSum[j] = 0;
        j++;
      }
      subArrSum[i] = 0;
    }
  }
 
  // Calculate total sum
  for(i = 0; i < subArrSum.Count; i++)
    totalSum += subArrSum[i];
 
  // Return totalSum
  return totalSum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = {3, 2, 3, 1, 4};
    int n = arr.Length;   
    Console.Write(findSubarraySum(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

输出：

41

时间复杂度： O(N * logN)

辅助空间： O(N)

方法2(基于哈希)的想法是制作一个空的哈希表。我们生成所有子数组。对于每个子数组，我们计算其总和以及哈希表中总和的增量计数。最后，我们将所有计数为1的总和相加。

C++

// C++ for finding sum of all unique subarray sum
#include 
using namespace std;
 
// function for finding grandSum
long long int findSubarraySum(int arr[], int n)
{
    int res = 0;
 
    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    unordered_map m;
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = i; j < n; j++) {
            sum += arr[j];
            m[sum]++;
        }
    }
 
    // Print all those sums that appear
    // once.
    for (auto x : m)
        if (x.second == 1)
            res += x.first;
 
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSubarraySum(arr, n);
    return 0;
}

Java

// Java for finding sum of all
// unique subarray sum
import java.util.*;
 
class GFG
{
 
// function for finding grandSum
static int findSubarraySum(int []arr, int n)
{
    int res = 0;
 
    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    HashMap m = new HashMap();
    for (int i = 0; i < n; i++)
    {
        int sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[j];
            if (m.containsKey(sum))
            {
                m.put(sum, m.get(sum) + 1);
            }
            else
            {
                m.put(sum, 1);
            }
        }
    }
 
    // Print all those sums that appear
    // once.
    for (Map.Entry x : m.entrySet())
        if (x.getValue() == 1)
            res += x.getKey();
 
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = arr.length;
    System.out.println(findSubarraySum(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 for finding sum of all
# unique subarray sum
 
# function for finding grandSum
def findSubarraySum(arr, n):
 
    res = 0
 
    # Go through all subarrays, compute sums
    # and count occurrences of sums.
    m = dict()
    for i in range(n):
        Sum = 0
        for j in range(i, n):
            Sum += arr[j]
            m[Sum] = m.get(Sum, 0) + 1
         
    # Print all those Sums that appear
    # once.
    for x in m:
        if m[x] == 1:
            res += x
 
    return res
 
# Driver code
arr = [3, 2, 3, 1, 4]
n = len(arr)
print(findSubarraySum(arr, n))
 
# This code is contributed by mohit kumar

C＃

// C# for finding sum of all
// unique subarray sum
using System;
using System.Collections.Generic;
     
class GFG
{
 
// function for finding grandSum
static int findSubarraySum(int []arr, int n)
{
    int res = 0;
 
    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    Dictionary m = new Dictionary();
    for (int i = 0; i < n; i++)
    {
        int sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[j];
            if (m.ContainsKey(sum))
            {
                m[sum] = m[sum] + 1;
            }
            else
            {
                m.Add(sum, 1);
            }
        }
    }
 
    // Print all those sums that appear
    // once.
    foreach(KeyValuePair x in m)
        if (x.Value == 1)
            res += x.Key;
 
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 3, 1, 4 };
    int n = arr.Length;
    Console.WriteLine(findSubarraySum(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

输出：

41

时间复杂度： O(N)

辅助空间： O(N)