给定二进制字符串的所有字符相同的子字符串的计数

给定只包含0和1 的二进制字符串str ，任务是找到分别只包含1和0的子字符串的数量，即所有字符相同。

例子：

Input: str = “011”
Output: 4
Explanation: 
Three sub-strings are “1“, “1”, “11” which have only 1 in them, and one substring is there which contains only “0”.

Input: str = “0000”
Output: 10
Explanation: 
There are no sub-strings having all ones in it.

朴素的方法：这个想法是生成给定字符串的所有可能的子字符串。对于每个子串检查如果字符串包含全部为1或全部为0。如果是，则计算该子字符串。上述操作后打印子串的计数。

时间复杂度：O(N ³ )
辅助空间：O(N)

高效的方法：这个想法是使用滑动窗口和两个指针方法的概念。以下是查找仅包含1s的子字符串的计数的步骤：

1. 初始化两个指针L和R并将它们初始化为0 。
2. 现在迭代给定的字符串并检查当前字符是否等于 1。如果是，则通过增加R的值来扩展窗口。
3. 如果当前字符为0 ，则窗口L 到 R – 1包含所有 1。
4. 将L 到 R – 1的子串的数量添加到((R – L) * (R – L + 1)) / 2的答案中，并增加R并将L重新初始化为 R 。
5. 重复这个过程，直到L 和 R相互交叉。
6. 打印第 4 步中所有子串的计数。
7. 要计算全为0的子字符串的数量，请翻转给定的字符串，即，全0将转换为1 ，反之亦然。
8. 对翻转后的字符串中的字符1重复上述步骤1 到步骤 4的步骤，以获取其中仅包含0的子字符串的计数并打印计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count number of
// sub-strings of a given binary
// string that contains only 1
int countSubAllOnes(string s)
{
    int l = 0, r = 0, ans = 0;
 
    // Iterate untill L and R cross
    // each other
    while (l <= r) {
 
        // Check if reached the end
        // of string
        if (r == s.length()) {
            ans += ((r - l) * (r - l + 1)) / 2;
            break;
        }
 
        // Check if encountered '1'
        // then extend window
        if (s[r] == '1')
            r++;
 
        // Check if encountered '0' then
        // add number of strings of
        // current window and change the
        // values for both l and r
        else {
 
            ans += ((r - l) * (r - l + 1)) / 2;
            l = r + 1;
            r++;
        }
    }
 
    // Return the answer
    return ans;
}
 
// Function to flip the bits of string
void flip(string& s)
{
 
    for (int i = 0; s[i]; i++) {
        if (s[i] == '1')
            s[i] = '0';
        else
            s[i] = '1';
    }
  cout<

---

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count number of
// sub-Strings of a given binary
// String that contains only 1
static int countSubAllOnes(String s)
{
    int l = 0, r = 0, ans = 0;
 
    // Iterate untill L and R cross
    // each other
    while (l <= r)
    {
         
        // Check if reached the end
        // of String
        if (r == s.length())
        {
            ans += ((r - l) * (r - l + 1)) / 2;
            break;
        }
 
        // Check if encountered '1'
        // then extend window
        if (s.charAt(r) == '1')
            r++;
 
        // Check if encountered '0' then
        // add number of Strings of
        // current window and change the
        // values for both l and r
        else
        {
            ans += ((r - l) * (r - l + 1)) / 2;
            l = r + 1;
            r++;
        }
    }
 
    // Return the answer
    return ans;
}
 
// Function to flip the bits of String
static String flip(char []s)
{
    for(int i = 0; i < s.length; i++)
    {
        if (s[i] == '1')
            s[i] = '0';
        else
            s[i] = '1';
    }
    return String.valueOf(s);
}
 
// Function to count number of
// sub-Strings of a given binary
// String that contains only 0s & 1s
static int countSubAllZerosOnes(String s)
{
 
    // count of subString
    // which contains only 1s
    int only_1s = countSubAllOnes(s);
 
    // Flip the character of String s
    // 0 to 1 and 1 to 0 to count the
    // subString with consecutive 0s
    s = flip(s.toCharArray());
 
    // count of subString
    // which contains only 0s
    int only_0s = countSubAllOnes(s);
 
    return only_0s + only_1s;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String str
    String s = "011";
 
    // Function call
    System.out.print(countSubAllZerosOnes(s) + "\n");
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 program for
# the above approach
 
# Function to count number of
# sub-strings of a given binary
# string that contains only 1
def countSubAllOnes(s):
   
    l, r, ans = 0, 0, 0
 
    # Iterate untill L and R cross
    # each other
    while (l <= r):
 
        # Check if reached the end
        # of string
        if (r == len(s)):
            ans += ((r - l) *
                    (r - l + 1)) // 2
            break
        
        # Check if encountered '1'
        # then extend window
        if (s[r] == '1'):
            r += 1
 
        # Check if encountered '0' then
        # add number of strings of
        # current window and change the
        # values for both l and r
        else :
            ans += ((r - l) *
                    (r - l + 1)) // 2
            l = r + 1
            r += 1
 
    # Return the answer
    return ans
 
# Function to flip the bits of string
def flip(s):
      
    arr = list(s)
    for i in range (len(s)):
        if (arr[i] == '1'):
            arr[i] = '0'
        else:
            arr[i] = '1'
    s = ''.join(arr)
    return s
 
# Function to count number of
# sub-strings of a given binary
# string that contains only 0s & 1s
def countSubAllZerosOnes(s):
 
    # count of substring
    # which contains only 1s
    only_1s = countSubAllOnes(s)
 
    # Flip the character of string s
    # 0 to 1 and 1 to 0 to count the
    # substring with consecutive 0s
    s = flip(s)
 
    # count of substring
    # which contains only 0s
    only_0s = countSubAllOnes(s)
 
    return only_0s + only_1s
 
# Driver Code
if __name__ == "__main__":
   
    # Given string str
    s  = "011"
 
    # Function Call
    print (countSubAllZerosOnes(s))
    
# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to count number of
// sub-Strings of a given binary
// String that contains only 1
static int countSubAllOnes(String s)
{
    int l = 0, r = 0, ans = 0;
 
    // Iterate untill L and R cross
    // each other
    while (l <= r)
    {
         
        // Check if reached the end
        // of String
        if (r == s.Length)
        {
            ans += ((r - l) * (r - l + 1)) / 2;
            break;
        }
 
        // Check if encountered '1'
        // then extend window
        if (s[r] == '1')
            r++;
 
        // Check if encountered '0' then
        // add number of Strings of
        // current window and change the
        // values for both l and r
        else
        {
            ans += ((r - l) * (r - l + 1)) / 2;
            l = r + 1;
            r++;
        }
    }
 
    // Return the answer
    return ans;
}
 
// Function to flip the bits of String
static String flip(char []s)
{
    for(int i = 0; i < s.Length; i++)
    {
        if (s[i] == '1')
            s[i] = '0';
        else
            s[i] = '1';
    }
    return String.Join("",s);
}
 
// Function to count number of
// sub-Strings of a given binary
// String that contains only 0s & 1s
static int countSubAllZerosOnes(String s)
{
 
    // count of subString
    // which contains only 1s
    int only_1s = countSubAllOnes(s);
 
    // Flip the character of String s
    // 0 to 1 and 1 to 0 to count the
    // subString with consecutive 0s
    s = flip(s.ToCharArray());
 
    // count of subString
    // which contains only 0s
    int only_0s = countSubAllOnes(s);
 
    return only_0s + only_1s;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String str
    String s = "011";
 
    // Function call
    Console.Write(countSubAllZerosOnes(s) + "\n");
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// Javascript program for the above approach
 
// Function to count number of
// sub-strings of a given binary
// string that contains only 1
function countSubAllOnes(s)
{
    var l = 0, r = 0, ans = 0;
 
    // Iterate untill L and R cross
    // each other
    while (l <= r) {
 
        // Check if reached the end
        // of string
        if (r == s.length) {
            ans += ((r - l) * (r - l + 1)) / 2;
            break;
        }
 
        // Check if encountered '1'
        // then extend window
        if (s[r] == '1')
            r++;
 
        // Check if encountered '0' then
        // add number of strings of
        // current window and change the
        // values for both l and r
        else {
 
            ans += ((r - l) * (r - l + 1)) / 2;
            l = r + 1;
            r++;
        }
    }
 
    // Return the answer
    return ans;
}
 
// Function to flip the bits of string
function flip(s)
{
 
    for (var i = 0; s[i]; i++) {
        if (s[i] == '1')
            s[i] = '0';
        else
            s[i] = '1';
    }
   
  return s;
}
 
// Function to count number of
// sub-strings of a given binary
// string that contains only 0s & 1s
function countSubAllZerosOnes(s)
{
 
    // count of substring
    // which contains only 1s
    var only_1s = countSubAllOnes(s);
 
    // Flip the character of string s
    // 0 to 1 and 1 to 0 to count the
    // substring with consecutive 0s
    s = flip(s.split(''));
 
    // count of substring
    // which contains only 0s
    var only_0s = countSubAllOnes(s);
 
    return only_0s + only_1s;
}
 
// Driver Code
 
// Given string str
var s = "011";
 
// Function Call
document.write( countSubAllZerosOnes(s));
 
// This code is contributed by famously.

输出：

4

时间复杂度： O(N)，其中 N 是给定字符串的长度。
辅助空间： O(1)

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