给定字符串S的有效括号长度为N ，偶数为K ，任务是找到长度为K的有效括号序列，该长度也是给定字符串。
注意：可以有多个有效序列，可以打印其中任何一个。
例子：
Input: S = “()()()”, K = 4
Output: ()()
Explanation:
The string “()()” is a subsequence of length 4 which is a valid parenthesis sequence.
Input: S = “()(())”, K = 6
Output: ()(())
Explanation:
The string “()(())” is a subsequence of length 6 which is a valid parenthesis sequence.
天真的方法：想法是生成给定字符串长度K的所有可能的子序列，并打印具有有效括号序列的任何字符串。
时间复杂度： O(2 N )
辅助空间： O(K)
高效方法：可以使用堆栈来优化上述方法。这个想法是遍历给定的字符串，当遇到一个开放的括号字符时，将其压入其他堆栈，然后从中弹出一个字符。相应地，每次弹出一个字符都增加计数器。请按照以下步骤解决问题：
- 创建一个堆栈和一个布尔数组，将其初始化为false 。
- 遍历给定的字符串，如果遇到开括号，则将该索引推入堆栈。
- 否则，如果遇到右括号：
- 从堆栈中弹出顶部元素
- 计数器加2
- 将弹出的标记和当前索引标记为true。
- 如果计数器超过K ，则终止。
- 遍历后，将所有字符从左到右附加在一起，标记为true。打印形成的结果字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
#define ll long long
using namespace std;
// Function to find the subsequence
// of length K forming valid sequence
string findString(string s, int k)
{
int n = s.length();
// Stores the resultant string
string ans = "";
stack st;
// Check whether character at
// index i is visited or not
vector vis(n, false);
int count = 0;
// Traverse the string
for (int i = 0; i < n; ++i) {
// Push index of open bracket
if (s[i] == '(') {
st.push(i);
}
// Pop and mark visited
if (count < k && s[i] == ')') {
vis[st.top()] = 1;
st.pop();
vis[i] = true;
// Increment count by 2
count += 2;
}
}
// Append the characters and create
// the resultant string
for (int i = 0; i < n; ++i) {
if (vis[i] == true) {
ans += s[i];
}
}
// Return the resultant string
return ans;
}
// Driver Code
int main()
{
string s = "()()()";
int K = 2;
// Function Call
cout << findString(s, K);
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the subsequence
// of length K forming valid sequence
static String findString(String s, int k)
{
int n = s.length();
// Stores the resultant String
String ans = " ";
Stack st = new Stack<>();
// Check whether character at
// index i is visited or not
boolean []vis = new boolean[n];
// Vector vis(n, false);
int count = 0;
// Traverse the String
for(int i = 0; i < n; ++i)
{
// Push index of open bracket
if (s.charAt(i) == '(')
{
st.add(i);
}
// Pop and mark visited
if (count < k && s.charAt(i) == ')')
{
vis[st.peek()] = true;
st.pop();
vis[i] = true;
// Increment count by 2
count += 2;
}
}
// Append the characters and create
// the resultant String
for(int i = 0; i < n; ++i)
{
if (vis[i] == true)
{
ans += s.charAt(i);
}
}
// Return the resultant String
return ans;
}
// Driver Code
public static void main(String[] args)
{
String s = "()()()";
int K = 2;
// Function call
System.out.print(findString(s, K));
}
}
// This code is contributed by gauravrajput1 Python3
# Python3 program for the above approach
# Function to find the subsequence
# of length K forming valid sequence
def findString(s, k):
n = len(s)
# Stores the resultant string
ans = ""
st = []
# Check whether character at
# index i is visited or not
vis = [False] * n
count = 0
# Traverse the string
for i in range(n):
# Push index of open bracket
if (s[i] == '('):
st.append(i)
# Pop and mark visited
if (count < k and s[i] == ')'):
vis[st[-1]] = 1
del st[-1]
vis[i] = True
# Increment count by 2
count += 2
# Append the characters and create
# the resultant string
for i in range(n):
if (vis[i] == True):
ans += s[i]
# Return the resultant string
return ans
# Driver Code
if __name__ == '__main__':
s = "()()()"
K = 2
# Function call
print(findString(s, K))
# This code is contributed by mohit kumar 29C#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the
// subsequence of length
// K forming valid sequence
static String findString(String s,
int k)
{
int n = s.Length;
// Stores the resultant String
String ans = " ";
Stack st = new Stack();
// Check whether character at
// index i is visited or not
bool []vis = new bool[n];
// List vis(n, false);
int count = 0;
// Traverse the String
for(int i = 0; i < n; ++i)
{
// Push index of open bracket
if (s[i] == '(')
{
st.Push(i);
}
// Pop and mark visited
if (count < k && s[i] == ')')
{
vis[st.Peek()] = true;
st.Pop();
vis[i] = true;
// Increment count by 2
count += 2;
}
}
// Append the characters and create
// the resultant String
for(int i = 0; i < n; ++i)
{
if (vis[i] == true)
{
ans += s[i];
}
}
// Return the resultant String
return ans;
}
// Driver Code
public static void Main(String[] args)
{
String s = "()()()";
int K = 2;
// Function call
Console.Write(findString(s, K));
}
}
// This code is contributed by Princi Singh 输出：
()
时间复杂度： O(N)
辅助空间： O(N)