打印给定范围内总和的所有子数组

给定一个正整数数组arr[]和定义范围[L, R]的两个整数L和R。任务是打印总和在L到R范围内的子数组。

例子：

Input: arr[] = {1, 4, 6}, L = 3, R = 8
Output: {1, 4}, {4}, {6}.
Explanation: All the possible subarrays are the following
{1] with sum 1. 
{1, 4} with sum 5. 
{1, 4, 6} with sum 11.
{4} with sum 4.
{4, 6} with sum 10. 
{6} with sum 6. 
Therefore, subarrays {1, 4}, {4}, {6} are having sum in range [3, 8].

Input: arr[] = {2, 3, 5, 8}, L = 4, R = 13
Output:  {2, 3}, {2, 3, 5}, {3, 5}, {5}, {5, 8}, {8}.

方法：这个问题可以通过暴力破解并使用两个循环检查每个可能的子数组来解决。下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find subarrays in given range
void subArraySum(int arr[], int n,
                 int leftsum, int rightsum)
{
    int curr_sum, i, j, res = 0;
 
    // Pick a starting point
    for (i = 0; i < n; i++) {
        curr_sum = arr[i];
 
        // Try all subarrays starting with 'i'
        for (j = i + 1; j <= n; j++) {
            if (curr_sum > leftsum
                && curr_sum < rightsum) {
                cout << "{ ";
 
                for (int k = i; k < j; k++)
                    cout << arr[k] << " ";
 
                cout << "}\n";
            }
            if (curr_sum > rightsum || j == n)
                break;
            curr_sum = curr_sum + arr[j];
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    int L = 10, R = 23;
 
    subArraySum(arr, N, L, R);
 
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG
{
   
    // Function to find subarrays in given range
    static void subArraySum(int arr[], int n, int leftsum,
                            int rightsum)
    {
        int curr_sum, i, j, res = 0;
 
        // Pick a starting point
        for (i = 0; i < n; i++) {
            curr_sum = arr[i];
 
            // Try all subarrays starting with 'i'
            for (j = i + 1; j <= n; j++) {
                if (curr_sum > leftsum
                    && curr_sum < rightsum) {
                    System.out.print("{ ");
 
                    for (int k = i; k < j; k++)
                        System.out.print(arr[k] + " ");
 
                    System.out.println("}");
                }
                if (curr_sum > rightsum || j == n)
                    break;
                curr_sum = curr_sum + arr[j];
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
        int N = arr.length;
 
        int L = 10, R = 23;
 
        subArraySum(arr, N, L, R);
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python program for above approach
 
# Function to find subarrays in given range
def subArraySum (arr, n, leftsum, rightsum):
    res = 0
 
    # Pick a starting point
    for i in range(n):
        curr_sum = arr[i]
 
        # Try all subarrays starting with 'i'
        for j in range(i + 1, n + 1):
            if (curr_sum > leftsum
                and curr_sum < rightsum):
                print("{ ", end="")
 
                for k in range(i, j):
                    print(arr[k], end=" ")
 
                print("}")
            if (curr_sum > rightsum or j == n):
                break
            curr_sum = curr_sum + arr[j]
         
# Driver Code
arr = [15, 2, 4, 8, 9, 5, 10, 23]
N = len(arr)
L = 10
R = 23
subArraySum(arr, N, L, R)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code for the above approach
using System;
 
class GFG
{
   
    // Function to find subarrays in given range
    static void subArraySum(int []arr, int n, int leftsum,
                            int rightsum)
    {
        int curr_sum, i, j, res = 0;
 
        // Pick a starting point
        for (i = 0; i < n; i++) {
            curr_sum = arr[i];
 
            // Try all subarrays starting with 'i'
            for (j = i + 1; j <= n; j++) {
                if (curr_sum > leftsum
                    && curr_sum < rightsum) {
                    Console.Write("{ ");
 
                    for (int k = i; k < j; k++)
                        Console.Write(arr[k] + " ");
 
                    Console.WriteLine("}");
                }
                if (curr_sum > rightsum || j == n)
                    break;
                curr_sum = curr_sum + arr[j];
            }
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int []arr = { 15, 2, 4, 8, 9, 5, 10, 23 };
        int N = arr.Length;
 
        int L = 10, R = 23;
 
        subArraySum(arr, N, L, R);
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for above approach
 
    // Function to find subarrays in given range
    const subArraySum = (arr, n, leftsum, rightsum) => {
        let curr_sum, i, j, res = 0;
 
        // Pick a starting point
        for (i = 0; i < n; i++) {
            curr_sum = arr[i];
 
            // Try all subarrays starting with 'i'
            for (j = i + 1; j <= n; j++) {
                if (curr_sum > leftsum
                    && curr_sum < rightsum) {
                    document.write("{ ");
 
                    for (let k = i; k < j; k++)
                        document.write(`${arr[k]} `);
 
                    document.write("}<br/>");
                }
                if (curr_sum > rightsum || j == n)
                    break;
                curr_sum = curr_sum + arr[j];
            }
        }
    }
 
    // Driver Code
    let arr = [15, 2, 4, 8, 9, 5, 10, 23];
    let N = arr.length;
    let L = 10, R = 23;
    subArraySum(arr, N, L, R);
 
    // This code is contributed by rakeshsahni
 

输出

{ 15 }
{ 15 2 }
{ 15 2 4 }
{ 2 4 8 }
{ 4 8 }
{ 4 8 9 }
{ 8 9 }
{ 8 9 5 }
{ 9 5 }
{ 5 10 }

时间复杂度： O(N^3)

辅助空间： O(1)