计算仅由一位数整数组成的子数组

给定一个由N 个正整数组成的数组arr[] ，任务是计算仅由一位数元素组成的子数组。

例子：

Input: arr[] = {0, 1, 14, 2, 5}
Output: 6
Explanation: All subarrays made of only single digit numbers are {{0}, {1}, {2}, {5}, {0, 1}, {2, 5}}. Therefore, the total count of subarrays is 6.

Input: arr[] ={12, 5, 14, 17}
Output: 1
Explanation: All subarrays made of only single digit numbers are {5}.
Therefore, the total count of subarrays is 1.

朴素的方法：最简单的方法是遍历数组并生成所有可能的子数组。对于每个子数组，检查其中的所有整数是否都是个位数整数。

时间复杂度： O(N ³ )
辅助空间： O(1)

有效的方法：为了优化上述方法，其思想是找到每个连续个位数整数块的大小，并按该长度的总子数组增加计数。请按照以下步骤解决问题：

• 初始化一个变量，比如res = 0和c = 0 ，以存储子数组的总数和一个子数组中个位数整数的总数。
• 遍历数组并执行以下操作：
  • 如果arr[i] < 10，将c的计数加 1，将res的计数加c。
  • 否则，分配c = 0。
• 最后，打印个位数整数子数组的总数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count of subarrays made
// up of single digit integers only
int singleDigitSubarrayCount(int arr[],
                             int N)
{
    // Stores count of subarrays
    int res = 0;
 
    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        if (arr[i] <= 9) {
 
            // Increment size of block by 1
            count++;
 
            // Increment res by count
            res += count;
        }
 
        else {
 
            // Assign count = 0
            count = 0;
        }
    }
 
    cout << res;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 0, 1, 14, 2, 5 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
    singleDigitSubarrayCount(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG
{
 
// Function to count of subarrays made
// up of single digit integers only
static void singleDigitSubarrayCount(int arr[],
                             int N)
{
   
    // Stores count of subarrays
    int res = 0;
 
    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
        if (arr[i] <= 9)
        {
 
            // Increment size of block by 1
            count++;
 
            // Increment res by count
            res += count;
        }
 
        else
        {
 
            // Assign count = 0
            count = 0;
        }
    }
    System.out.print(res);
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given array
    int arr[] = { 0, 1, 14, 2, 5 };
 
    // Size of the array
    int N = arr.length;
    singleDigitSubarrayCount(arr, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to count of subarrays made
# up of single digit integers only
def singleDigitSubarrayCount(arr, N):
     
    # Stores count of subarrays
    res = 0
 
    # Stores the count of consecutive
    # single digit numbers in the array
    count = 0
 
    # Traverse the array
    for i in range(N):
        if (arr[i] <= 9):
 
            # Increment size of block by 1
            count += 1
 
            # Increment res by count
            res += count
        else:
            # Assign count = 0
            count = 0
    print (res)
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [0, 1, 14, 2, 5]
 
    # Size of the array
    N = len(arr)
    singleDigitSubarrayCount(arr, N)
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to count of subarrays made
// up of single digit integers only
static void singleDigitSubarrayCount(int[] arr,
                             int N)
{
   
    // Stores count of subarrays
    int res = 0;
 
    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
        if (arr[i] <= 9)
        {
 
            // Increment size of block by 1
            count++;
 
            // Increment res by count
            res += count;
        }
        else
        {
 
            // Assign count = 0
            count = 0;
        }
    }
    Console.Write(res);
}
 
// Driver Code
public static void Main(string[] args)
{
    // Given array
    int[] arr = { 0, 1, 14, 2, 5 };
 
    // Size of the array
    int N = arr.Length;
    singleDigitSubarrayCount(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Javascript

 
// Javascript program for the above approach
 
// Function to count of subarrays made
// up of single digit integers only
function singleDigitSubarrayCount(arr, N)
{
     
    // Stores count of subarrays
    let res = 0;
 
    // Stores the count of consecutive
    // single digit numbers in the array
    let count = 0;
 
    // Traverse the array
    for(let i = 0; i < N; i++)
    {
        if (arr[i] <= 9)
        {
             
            // Increment size of block by 1
            count++;
 
            // Increment res by count
            res += count;
        }
        else
        {
             
            // Assign count = 0
            count = 0;
        }
    }
    document.write(res);
}
 
// Driver Code
 
// Given array
let arr = [ 0, 1, 14, 2, 5 ];
 
// Size of the array
let N = arr.length;
 
singleDigitSubarrayCount(arr, N);
 
// This code is contributed by Manoj.
 

输出：

6

时间复杂度： O(N)
辅助空间： O(1)

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