平衡括号子串的数量
给定一个由“ ( )和“ ) 组成的平衡括号字符串。任务是找到给定字符串 中平衡括号子串的数量
例:
输入:str =()())()” T3】输出: 6 ()、()、()、()、()、()、())() T7】输入:str =(()))()” T10】输出: 4 )、()、()、()、()、()、())()
方法: 让我们假设，每当我们遇到左括号时，深度增加 1，当遇到右括号时，深度减少 1。每当我们遇到结束括号时，将我们需要的答案增加 1，然后在这个深度用已经形成的平衡子串增加我们需要的答案。 以下是上述办法的实施:
C++
// CPP program to find number of
// balanced parenthesis sub strings
#include <bits/stdc++.h>
using namespace std;
// Function to find number of
// balanced parenthesis sub strings
int Balanced_Substring(string str, int n)
{
// To store required answer
int ans = 0;
// Vector to stores the number of
// balanced brackets at each depth.
vector<int> arr(n / 2 + 1, 0);
// d stores checks the depth of our sequence
// For example level of () is 1
// and that of (()) is 2.
int d = 0;
for (int i = 0; i < n; i++) {
// If open bracket
// increase depth
if (str[i] == '(')
d++;
// If closing bracket
else {
if (d == 1) {
for (int j = 2; j <= n / 2 + 1 && arr[j] != 0; j++)
arr[j] = 0;
}
++ans;
ans += arr[d];
arr[d]++;
d--;
}
}
// Return the required answer
return ans;
}
// Driver code
int main()
{
string str = "()()()";
int n = str.size();
// Function call
cout << Balanced_Substring(str, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find number of
// balanced parenthesis sub strings
class GFG {
// Function to find number of
// balanced parenthesis sub strings
public static int Balanced_Substring(String str,
int n)
{
// To store required answer
int ans = 0;
// Vector to stores the number of
// balanced brackets at each depth.
int[] arr = new int[n / 2 + 1];
// d stores checks the depth of our sequence
// For example level of () is 1
// and that of (()) is 2.
int d = 0;
for (int i = 0; i < n; i++) {
// If open bracket
// increase depth
if (str.charAt(i) == '(')
d++;
// If closing bracket
else {
if (d == 1) {
for (int j = 2; j <= n / 2 + 1 && arr[j] != 0; j++)
arr[j] = 0;
}
++ans;
ans += arr[d];
arr[d]++;
d--;
}
}
// Return the required answer
return ans;
}
// Driver code
public static void main(String[] args)
{
String str = "()()()";
int n = str.length();
// Function call
System.out.println(Balanced_Substring(str, n));
}
}
// This code is contributed by
// sanjeev2552
Python 3
# Python3 program to find number of
# balanced parenthesis sub strings
# Function to find number of
# balanced parenthesis sub strings
def Balanced_Substring(s, n):
# To store required answer
ans = 0;
# Vector to stores the number of
# balanced brackets at each depth.
arr = [0] * (int(n / 2) + 1);
# d stores checks the depth of our sequence
# For example level of () is 1
# and that of (()) is 2.
d = 0;
for i in range(n):
# If open bracket
# increase depth
if (s[i] == '('):
d += 1;
# If closing bracket
else:
if (d == 1):
j = 2
while (j <= n//2 + 1 and arr[j] != 0):
arr[j] = 0
ans += 1;
ans += arr[d];
arr[d] += 1;
d -= 1;
# Return the required answer
return ans;
# Driver code
s = "()()()";
n = len(s);
# Function call
print(Balanced_Substring(s, n));
# This code contributed by Rajput-Ji
C
// C# program to find number of
// balanced parenthesis sub strings
using System;
class GFG {
// Function to find number of
// balanced parenthesis sub strings
public static int Balanced_Substring(String str,
int n)
{
// To store required answer
int ans = 0;
// Vector to stores the number of
// balanced brackets at each depth.
int[] arr = new int[n / 2 + 1];
// d stores checks the depth of our sequence
// For example level of () is 1
// and that of (()) is 2.
int d = 0;
for (int i = 0; i < n; i++) {
// If open bracket
// increase depth
if (str[i] == '(')
d++;
// If closing bracket
else {
if (d == 1) {
for (int j = 2; j <= n / 2 + 1 && arr[j] != 0; j++)
arr[j] = 0;
}
++ans;
ans += arr[d];
arr[d]++;
d--;
}
}
// Return the required answer
return ans;
}
// Driver code
public static void Main(String[] args)
{
String str = "()()()";
int n = str.Length;
// Function call
Console.WriteLine(Balanced_Substring(str, n));
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to find number of
// balanced parenthesis sub strings
// Function to find number of
// balanced parenthesis sub strings
function Balanced_Substring(str, n)
{
// To store required answer
let ans = 0;
// Vector to stores the number of
// balanced brackets at each depth.
let arr = new Array(n / 2 + 1).fill(0);
// d stores checks the depth of our sequence
// For example level of () is 1
// and that of (()) is 2.
let d = 0;
for (let i = 0; i < n; i++) {
// If open bracket
// increase depth
if (str[i] == '(')
d++;
// If closing bracket
else {
if (d == 1) {
for (let j = 2; j <= parseInt(n / 2) +
1 && arr[j] != 0; j++)
arr[j] = 0;
}
++ans;
ans += arr[d];
arr[d]++;
d--;
}
}
// Return the required answer
return ans;
}
// Driver code
let str = "()()()";
let n = str.length;
// Function call
document.write(Balanced_Substring(str, n));
</script>
Output:
6
时间复杂度: O(N)
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