所有大小为 K 的子数组的总和

给定一个数组arr[]和一个整数K ，任务是计算所有大小为 K 的子数组的总和。

例子：

Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 3 
Output: 6 9 12 15 
Explanation: 
All subarrays of size k and their sum: 
Subarray 1: {1, 2, 3} = 1 + 2 + 3 = 6 
Subarray 2: {2, 3, 4} = 2 + 3 + 4 = 9 
Subarray 3: {3, 4, 5} = 3 + 4 + 5 = 12 
Subarray 4: {4, 5, 6} = 4 + 5 + 6 = 15

Input: arr[] = {1, -2, 3, -4, 5, 6}, K = 2 
Output: -1, 1, -1, 1, 11 
Explanation: 
All subarrays of size K and their sum: 
Subarray 1: {1, -2} = 1 – 2 = -1 
Subarray 2: {-2, 3} = -2 + 3 = -1 
Subarray 3: {3, 4} = 3 – 4 = -1 
Subarray 4: {-4, 5} = -4 + 5 = 1 
Subarray 5: {5, 6} = 5 + 6 = 11 

朴素的方法：朴素的方法是生成大小为K 的所有子数组，并使用迭代找到每个子数组的总和。

下面是上述方法的实现：

C++

// C++ implementation to find the sum
// of all subarrays of size K
 
#include 
using namespace std;
 
// Function to find the sum of
// all subarrays of size K
int calcSum(int arr[], int n, int k)
{
 
    // Loop to consider every
    // subarray of size K
    for (int i = 0; i <= n - k; i++) {
         
        // Initialize sum = 0
        int sum = 0;
 
        // Calculate sum of all elements
        // of current subarray
        for (int j = i; j < k + i; j++)
            sum += arr[j];
 
        // Print sum of each subarray
        cout << sum << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    // Function Call
    calcSum(arr, n, k);
 
    return 0;
}

Java

// Java implementation to find the sum
// of all subarrays of size K
class GFG{
  
// Function to find the sum of
// all subarrays of size K
static void calcSum(int arr[], int n, int k)
{
  
    // Loop to consider every
    // subarray of size K
    for (int i = 0; i <= n - k; i++) {
          
        // Initialize sum = 0
        int sum = 0;
  
        // Calculate sum of all elements
        // of current subarray
        for (int j = i; j < k + i; j++)
            sum += arr[j];
  
        // Print sum of each subarray
        System.out.print(sum+ " ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = arr.length;
    int k = 3;
  
    // Function Call
    calcSum(arr, n, k);
}
}
 
// This code is contributed by Rajput-Ji

C#

// C# implementation to find the sum
// of all subarrays of size K
using System;
 
class GFG 
{
   
    // Function to find the sum of
    // all subarrays of size K
    static  void calcSum(int[] arr, int n, int k)
    {
     
        // Loop to consider every
        // subarray of size K
        for (int i = 0; i <= n - k; i++) {
             
            // Initialize sum = 0
            int sum = 0;
     
            // Calculate sum of all elements
            // of current subarray
            for (int j = i; j < k + i; j++)
                sum += arr[j];
     
            // Print sum of each subarray
            Console.Write(sum + " ");
        }
    }
     
    // Driver Code
    static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
        int n = arr.Length;
        int k = 3;
     
        // Function Call
        calcSum(arr, n, k);
     
    }
}
 
// This code is contributed by shubhamsingh10

Python3

# Python3 implementation to find the sum
# of all subarrays of size K
 
# Function to find the sum of
# all subarrays of size K
def calcSum(arr, n, k):
 
    # Loop to consider every
    # subarray of size K
    for i in range(n - k + 1):
         
        # Initialize sum = 0
        sum = 0
 
        # Calculate sum of all elements
        # of current subarray
        for j in range(i, k + i):
            sum += arr[j]
 
        # Prsum of each subarray
        print(sum, end=" ")
 
# Driver Code
arr=[1, 2, 3, 4, 5, 6]
n = len(arr)
k = 3
 
# Function Call
calcSum(arr, n, k)
 
# This code is contributed by mohit kumar 29

Javascript

 
// JavaScript implementation to find the sum
// of all subarrays of size K
 
// Function to find the sum of
// all subarrays of size K
function calcSum(arr, n, k)
{
 
    // Loop to consider every
    // subarray of size K
    for (var i = 0; i <= n - k; i++) {
         
        // Initialize sum = 0
        var sum = 0;
 
        // Calculate sum of all elements
        // of current subarray
        for (var j = i; j < k + i; j++)
            sum += arr[j];
 
        // Print sum of each subarray
        document.write(sum + " ");
    }
}
 
// Driver Code
var arr = [ 1, 2, 3, 4, 5, 6 ];
var n = arr.length;
var k = 3;
 
// Function Call
calcSum(arr, n, k);
 
 

输出：

6 9 12 15

性能分析：

• 时间复杂度：如上述方法，有两个循环，其中第一个循环运行 (N – K) 次，第二个循环运行 K 次。因此时间复杂度将为O(N*K) 。
• 辅助空间复杂度：与上述方法一样，没有使用额外的空间。因此辅助空间复杂度将为O(1) 。

有效的方法：使用滑动窗口这个想法是使用滑动窗口方法来找到数组中所有可能子数组的总和。

• 对于 [0, K] 范围内的每个大小，找到大小为 K 的第一个窗口的总和并将其存储在一个数组中。
• 然后对于 [K, N] 范围内的每个大小，添加下一个有助于滑动窗口的元素并减去从窗口弹出的元素。

// Adding the element which
// adds into the new window
sum = sum + arr[j]

// Subtracting the element which
// pops out from the window
sum = sum - arr[j-k]

where sum is the variable to store the result
      arr is the given array
      j is the loop variable in range [K, N]

下面是上述方法的实现：

C++

// C++ implementation to find the sum
// of all subarrays of size K
 
#include 
using namespace std;
 
// Function to find the sum of
// all subarrays of size K
int calcSum(int arr[], int n, int k)
{
    // Initialize sum = 0
    int sum = 0;
 
    // Consider first subarray of size k
    // Store the sum of elements
    for (int i = 0; i < k; i++)
        sum += arr[i];
 
    // Print the current sum
    cout << sum << " ";
 
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (int i = k; i < n; i++) {
         
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
         
        // Print the sum of subarray
        cout << sum << " ";
    }
}
 
// Drivers Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
     
    // Function Call
    calcSum(arr, n, k);
 
    return 0;
}

Java

// Java implementation to find the sum
// of all subarrays of size K
class GFG{
 
// Function to find the sum of
// all subarrays of size K
static void calcSum(int arr[], int n, int k)
{
    // Initialize sum = 0
    int sum = 0;
 
    // Consider first subarray of size k
    // Store the sum of elements
    for (int i = 0; i < k; i++)
        sum += arr[i];
 
    // Print the current sum
    System.out.print(sum+ " ");
 
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (int i = k; i < n; i++) {
         
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
         
        // Print the sum of subarray
        System.out.print(sum+ " ");
    }
}
 
// Drivers Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = arr.length;
    int k = 3;
     
    // Function Call
    calcSum(arr, n, k);
}
}
 
// This code is contributed by sapnasingh4991

蟒蛇3

# Python3 implementation to find the sum
# of all subarrays of size K
 
# Function to find the sum of
# all subarrays of size K
def calcSum(arr, n, k):
 
    # Initialize sum = 0
    sum = 0
 
    # Consider first subarray of size k
    # Store the sum of elements
    for i in range( k):
        sum += arr[i]
 
    # Print the current sum
    print( sum ,end= " ")
 
    # Consider every subarray of size k
    # Remove first element and add current
    # element to the window
    for i in range(k,n):
         
        # Add the element which enters
        # into the window and substract
        # the element which pops out from
        # the window of the size K
        sum = (sum - arr[i - k]) + arr[i]
         
        # Print the sum of subarray
        print( sum ,end=" ")
 
# Drivers Code
if __name__ == "__main__":
 
    arr = [ 1, 2, 3, 4, 5, 6 ]
    n = len(arr)
    k = 3
     
    # Function Call
    calcSum(arr, n, k)
 
# This code is contributed by chitranayal

C＃

// C# implementation to find the sum
// of all subarrays of size K
using System;
 
class GFG{
  
// Function to find the sum of
// all subarrays of size K
static void calcSum(int []arr, int n, int k)
{
    // Initialize sum = 0
    int sum = 0;
  
    // Consider first subarray of size k
    // Store the sum of elements
    for (int i = 0; i < k; i++)
        sum += arr[i];
  
    // Print the current sum
    Console.Write(sum+ " ");
  
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (int i = k; i < n; i++) {
          
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
          
        // Print the sum of subarray
        Console.Write(sum + " ");
    }
}
  
// Drivers Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    int n = arr.Length;
    int k = 3;
      
    // Function Call
    calcSum(arr, n, k);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation to find the sum
// of all subarrays of size K
 
// Function to find the sum of
// all subarrays of size K
function calcSum(arr, n, k)
{
 
    // Initialize sum = 0
    var sum = 0;
 
    // Consider first subarray of size k
    // Store the sum of elements
    for (var i = 0; i < k; i++)
        sum += arr[i];
 
    // Print the current sum
    document.write( sum + " ");
 
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (var i = k; i < n; i++) {
         
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
         
        // Print the sum of subarray
        document.write( sum + " ");
    }
}
 
// Drivers Code
var arr = [ 1, 2, 3, 4, 5, 6 ];
var n = arr.length;
var k = 3;
 
// Function Call
calcSum(arr, n, k);
 
// This code is contributed by noob2000.

输出：

6 9 12 15

性能分析：

• 时间复杂度：与上述方法一样，有一个循环需要 O(N) 时间。因此时间复杂度将为O(N) 。
• 辅助空间复杂度：与上述方法一样，没有使用额外的空间。因此辅助空间复杂度将为O(1) 。

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