给定二进制字符串str仅包含0和1 ，任务是查找分别仅包含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.
天真的方法：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。完成上述操作后，打印子字符串的计数。
时间复杂度：O(N 3 )
辅助空间：O(N)
高效的方法：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1的子字符串计数的步骤：
- 初始化两个指针，即L和R ，并将它们初始化为0 。
- 现在，在给定的字符串进行迭代，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。
- 如果当前字符为0 ，则窗口L到R – 1包含全1。
- 将L的子字符串数加到R – 1到答案为((R – L)*(R – L + 1))/ 2上，并递增R并将L重新初始化为R。
- 重复该过程，直到L和R彼此交叉。
- 在步骤4中打印所有子字符串的计数。
- 要计数全为0的子字符串的数量，请翻转给定的字符串，即，全0将被转换为1 ，反之亦然。
- 对翻转后的字符串中的字符1重复从步骤1到步骤4的上述步骤，以获取其中仅包含0的子字符串的计数并打印该计数。
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_ranjanPython3
# 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 ChitranayalC#
// 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)