给定一个二进制字符串，计算以 1 开头和结尾的子字符串的数量。

给定一个二进制字符串，计算以 1 开头和结尾的子字符串的数量。例如，如果输入字符串是“00100101”，则有三个子字符串“1001”、“100101”和“101”。
来源：亚马逊面试经历 |设置 162
难度级别：新手

一个简单的解决方案是运行两个循环。外循环选择每个 1 作为起点，内循环搜索结束 1 并在找到 1 时递增计数。

C++

// A simple C++ program to count number of
// substrings starting and ending with 1
#include
  
using namespace std;
 
int countSubStr(char str[])
{
int res = 0; // Initialize result
 
// Pick a starting point
for (int i=0; str[i] !='\0'; i++)
{
        if (str[i] == '1')
        {
            // Search for all possible ending point
            for (int j=i+1; str[j] !='\0'; j++)
            if (str[j] == '1')
                res++;
        }
}
return res;
}
 
// Driver program to test above function
int main()
{
char str[] = "00100101";
cout << countSubStr(str);
return 0;
}

Java

// A simple C++ program to count number of
//substrings starting and ending with 1
 
class CountSubString
{
    int countSubStr(char str[],int n)
    {
        int res = 0;  // Initialize result
 
        // Pick a starting point
        for (int i = 0; i

---

Python3

# A simple Python 3 program to count number of
# substrings starting and ending with 1
 
def countSubStr(st, n) :
     
    # Initialize result
    res = 0  
  
   # Pick a starting point
    for i in range(0, n) :
        if (st[i] == '1') :
 
            # Search for all possible ending point
            for j in range(i+1, n) :
                if (st[j] == '1') :
                    res = res + 1
         
    return res
     
  
# Driver program to test above function
st = "00100101";
list(st)
n= len(st)
print(countSubStr(st, n), end="")
 
 
# This code is contributed
# by Nikita Tiwari.

C#

// A simple C# program to count number of
// substrings starting and ending with 1
using System;
 
class GFG
{
public virtual int countSubStr(char[] str,
                               int n)
{
    int res = 0; // Initialize result
 
    // Pick a starting point
    for (int i = 0; i < n; i++)
    {
        if (str[i] == '1')
        {
            // Search for all possible
            // ending point
            for (int j = i + 1; j < n; j++)
            {
                if (str[j] == '1')
                {
                    res++;
                }
            }
        }
    }
    return res;
}
 
// Driver Code
public static void Main(string[] args)
{
    GFG count = new GFG();
    string s = "00100101";
    char[] str = s.ToCharArray();
    int n = str.Length;
    Console.WriteLine(count.countSubStr(str,n));
}
}
 
// This code is contributed by Shrikant13

Javascript

 
// A simple javascript program to count number of
// substrings starting and ending with 1
     
    function countSubStr(str,n)
    {
        let res = 0;  // Initialize result
        // Pick a starting point
        for (let i = 0; i<n; i++)
        {
            if (str[i] == '1')
            {
                // Search for all possible ending point
                for (let j = i + 1; j< n; j++)
                {
                    if (str[j] == '1')
                        res++;
                }
            }
        }
        return res;
    }
     
    // Driver program to test the above function
    let string = "00100101";
    let n=string.length;
    document.write(countSubStr(string,n));
     
     
    // This code is contributed by rag2127
     

输出：

3

上述解决方案的时间复杂度为 O(n 2 )。我们可以使用输入字符串的单次遍历在 O(n) 中找到计数。以下是步骤。
a) 计算 1 的个数。设 1 的计数为 m。
b) 返回 m(m-1)/2
这个想法是计算可能的 1 对的总数。


C++

// A O(n) C++ program to count number of
// substrings starting and ending with 1
#include
 
using namespace std;
 
int countSubStr(char str[])
{
   int m = 0; // Count of 1's in input string
 
   // Traverse input string and count of 1's in it
   for (int i=0; str[i] !='\0'; i++)
   {
        if (str[i] == '1')
           m++;
   }
 
   // Return count of possible pairs among m 1's
   return m*(m-1)/2;
}
 
// Driver program to test above function
int main()
{
  char str[] = "00100101";
  cout << countSubStr(str);
  return 0;
}

Java

// A O(n) C++ program to count number of substrings
//starting and ending with 1
 
class CountSubString
{
    int countSubStr(char str[], int n)
    {
        int m = 0; // Count of 1's in input string
 
        // Traverse input string and count of 1's in it
        for (int i = 0; i < n; i++)
        {
            if (str[i] == '1')
                m++;
        }
 
        // Return count of possible pairs among m 1's
        return m * (m - 1) / 2;
    }
 
    // Driver program to test the above function
    public static void main(String[] args)
    {
        CountSubString count = new CountSubString();
        String string = "00100101";
        char str[] = string.toCharArray();
        int n = str.length;
        System.out.println(count.countSubStr(str, n));
    }
}

Python3

# A Python3 program to count number of
# substrings starting and ending with 1
 
def countSubStr(st, n) :
 
    # Count of 1's in input string
    m = 0 
  
    # Traverse input string and
    # count of 1's in it
    for i in range(0, n) :
        if (st[i] == '1') :
            m = m + 1
         
    # Return count of possible
    # pairs among m 1's
    return m * (m - 1) // 2
    
  
# Driver program to test above function
st = "00100101";
list(st)
n= len(st)
print(countSubStr(st, n), end="")
 
 
# This code is contributed
# by Nikita Tiwari.

C＃

// A O(n) C# program to count
// number of substrings starting
// and ending with 1
using System;
 
class GFG
{
int countSubStr(char []str, int n)
{
    int m = 0; // Count of 1's in
               // input string
 
    // Traverse input string and
    // count of 1's in it
    for (int i = 0; i < n; i++)
    {
        if (str[i] == '1')
            m++;
    }
 
    // Return count of possible
    // pairs among m 1's
    return m * (m - 1) / 2;
}
 
// Driver Code
public static void Main(String[] args)
{
    GFG count = new GFG();
    String strings = "00100101";
    char []str = strings.ToCharArray();
    int n = str.Length;
    Console.Write(count.countSubStr(str, n));
}
}
 
// This code is contributed by princiraj

Javascript

// A O(n) javascript program to count number of substrings
//starting and ending with 1
 
    function countSubStr(str,n)
    {
        let m = 0; // Count of 1's in input string
  
        // Traverse input string and count of 1's in it
        for (let i = 0; i < n; i++)
        {
            if (str[i] == '1')
                m++;
        }
  
        // Return count of possible pairs among m 1's
        return m * Math.floor((m - 1) / 2);
    }
     
    // Driver program to test the above function
    let str = "00100101";
    let n = str.length;
    document.write(countSubStr(str, n));
     
    //  This code is contributed by avanitrachhadiya2155

输出：


3