给定二进制字符串str仅为0和1 。任务是计算字符串str的子字符串总数，以使每个子字符串中的连续0和1相等。
例子
Input: str = “010011”
Output: 4
Explanation:
The substrings with consecutive 0’s and 1’s are “01”, “10”, “0011”, “01”. Hence, the count is 4.
Note:
The two “01” are at different positions: [0, 1] and [3, 4].
“010011” has the same number of 0’s and 1’s but they are not consecutive.
Input: str = “0001110010”
Output: 7
Explanation:
The substrings with consecutive 0’s and 1’s are “000111”, “0011”, “01”, “1100”, “10”, “01”, “10”.
方法：
- 从字符串开始算起连续的0(或1)数。
- 然后从字符串str中0(或1)的计数结束的位置开始计数连续的1(或0)的数目。
- 具有连续0和1的子字符串的总数是在上述两个步骤中找到的连续0和1的计数的最小值。
- 重复上述步骤，直到字符串str结束。
下面是上述方法的实现：
C++
// C++ implementation of the
// above approach
#include
using namespace std;
// Function to find the count
// of substrings with equal no.
// of consecutive 0's and 1's
int countSubstring(string& S, int& n)
{
// To store the total count
// of substrings
int ans = 0;
int i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive
// 0's & 1's
int cnt0 = 0, cnt1 = 0;
// Counting subarrays of
// type "01"
if (S[i] == '0') {
// Count the consecutive
// 0's
while (i < n && S[i] == '0') {
cnt0++;
i++;
}
// If consecutive 0's
// ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S[j] == '1') {
cnt1++;
j++;
}
}
// Counting subarrays of
// type "10"
else {
// Count consecutive 1's
while (i < n && S[i] == '1') {
cnt1++;
i++;
}
// If consecutive 1's
// ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S[j] == '0') {
cnt0++;
j++;
}
}
// Update the total count
// of substrings with
// minimum of (cnt0, cnt1)
ans += min(cnt0, cnt1);
}
// Return answer
return ans;
}
// Driver code
int main()
{
string S = "0001110010";
int n = S.length();
// Function to print the
// count of substrings
cout << countSubstring(S, n);
return 0;
} Java
// Java implementation of the
// above approach
class GFG{
// Function to find the count
// of substrings with equal no.
// of consecutive 0's and 1's
static int countSubstring(String S, int n)
{
// To store the total count
// of substrings
int ans = 0;
int i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive
// 0's & 1's
int cnt0 = 0, cnt1 = 0;
// Counting subarrays of
// type "01"
if (S.charAt(i) == '0') {
// Count the consecutive
// 0's
while (i < n && S.charAt(i) == '0') {
cnt0++;
i++;
}
// If consecutive 0's
// ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S.charAt(j) == '1') {
cnt1++;
j++;
}
}
// Counting subarrays of
// type "10"
else {
// Count consecutive 1's
while (i < n && S.charAt(i) == '1') {
cnt1++;
i++;
}
// If consecutive 1's
// ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S.charAt(j) == '0') {
cnt0++;
j++;
}
}
// Update the total count
// of substrings with
// minimum of (cnt0, cnt1)
ans += Math.min(cnt0, cnt1);
}
// Return answer
return ans;
}
// Driver code
static public void main(String args[])
{
String S = "0001110010";
int n = S.length();
// Function to print the
// count of substrings
System.out.println(countSubstring(S, n));
}
}
// This code is contributed by Yash_RPython3
# Python3 implementation of the
# above approach
# Function to find the count
# of substrings with equal no.
# of consecutive 0's and 1's
def countSubstring(S, n) :
# To store the total count
# of substrings
ans = 0;
i = 0;
# Traversing the string
while (i < n) :
# Count of consecutive
# 0's & 1's
cnt0 = 0; cnt1 = 0;
# Counting subarrays of
# type "01"
if (S[i] == '0') :
# Count the consecutive
# 0's
while (i < n and S[i] == '0') :
cnt0 += 1;
i += 1;
# If consecutive 0's
# ends then check for
# consecutive 1's
j = i;
# Counting consecutive 1's
while (j < n and S[j] == '1') :
cnt1 += 1;
j += 1;
# Counting subarrays of
# type "10"
else :
# Count consecutive 1's
while (i < n and S[i] == '1') :
cnt1 += 1;
i += 1;
# If consecutive 1's
# ends then check for
# consecutive 0's
j = i;
# Count consecutive 0's
while (j < n and S[j] == '0') :
cnt0 += 1;
j += 1;
# Update the total count
# of substrings with
# minimum of (cnt0, cnt1)
ans += min(cnt0, cnt1);
# Return answer
return ans;
# Driver code
if __name__ == "__main__" :
S = "0001110010";
n = len(S);
# Function to print the
# count of substrings
print(countSubstring(S, n));
# This code is contributed by Yash_RC#
// C# implementation of the
// above approach
using System;
class GFG{
// Function to find the count
// of substrings with equal no.
// of consecutive 0's and 1's
static int countSubstring(string S, int n)
{
// To store the total count
// of substrings
int ans = 0;
int i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive
// 0's & 1's
int cnt0 = 0, cnt1 = 0;
// Counting subarrays of
// type "01"
if (S[i] == '0') {
// Count the consecutive
// 0's
while (i < n && S[i] == '0') {
cnt0++;
i++;
}
// If consecutive 0's
// ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S[j] == '1') {
cnt1++;
j++;
}
}
// Counting subarrays of
// type "10"
else {
// Count consecutive 1's
while (i < n && S[i] == '1') {
cnt1++;
i++;
}
// If consecutive 1's
// ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S[j] == '0') {
cnt0++;
j++;
}
}
// Update the total count
// of substrings with
// minimum of (cnt0, cnt1)
ans += Math.Min(cnt0, cnt1);
}
// Return answer
return ans;
}
// Driver code
static public void Main ()
{
string S = "0001110010";
int n = S.Length;
// Function to print the
// count of substrings
Console.WriteLine(countSubstring(S, n));
}
}
// This code is contributed by Yash_R输出：
7
时间复杂度： O(N)，其中N =字符串的长度。