将给出一个表达式，该表达式可以包含开括号和闭括号以及可选的一些字符，字符串没有其他运算符。我们需要删除最小括号，以使输入字符串有效。如果可能有多个有效输出，请删除相同数量的括号，然后打印所有此类输出。
例子：
Input : str = “()())()” -
Output : ()()() (())()
There are two possible solutions
"()()()" and "(())()"
Input : str = (v)())()
Output : (v)()() (v())()
当我们需要生成所有可能的输出时，我们将通过移除一个左括号或右括号来回溯所有状态，并检查它们是否有效(如果无效)，然后将移除的括号添加回去并进入下一个状态。我们将使用BFS来遍历各州，使用BFS将确保删除最少数量的括号，因为我们逐级遍历各个州，每个级别对应于一个额外的括号删除。除此之外，BFS不涉及任何递归，因此还节省了传递参数的开销。
下面的代码有一个isValidString方法来检查字符串的有效性，它在每个索引处都计算开括号和闭括号，而忽略非括号字符。如果在任何时候，右括号的计数变得大于开括号，那么我们返回false，否则我们将继续更新count变量。
C++
/* C/C++ program to remove invalid parenthesis */
#include
using namespace std;
// method checks if character is parenthesis(open
// or closed)
bool isParenthesis(char c)
{
return ((c == '(') || (c == ')'));
}
// method returns true if string contains valid
// parenthesis
bool isValidString(string str)
{
int cnt = 0;
for (int i = 0; i < str.length(); i++)
{
if (str[i] == '(')
cnt++;
else if (str[i] == ')')
cnt--;
if (cnt < 0)
return false;
}
return (cnt == 0);
}
// method to remove invalid parenthesis
void removeInvalidParenthesis(string str)
{
if (str.empty())
return ;
// visit set to ignore already visited string
set visit;
// queue to maintain BFS
queue q;
string temp;
bool level;
// pushing given string as starting node into queue
q.push(str);
visit.insert(str);
while (!q.empty())
{
str = q.front(); q.pop();
if (isValidString(str))
{
cout << str << endl;
// If answer is found, make level true
// so that valid string of only that level
// are processed.
level = true;
}
if (level)
continue;
for (int i = 0; i < str.length(); i++)
{
if (!isParenthesis(str[i]))
continue;
// Removing parenthesis from str and
// pushing into queue,if not visited already
temp = str.substr(0, i) + str.substr(i + 1);
if (visit.find(temp) == visit.end())
{
q.push(temp);
visit.insert(temp);
}
}
}
}
// Driver code to check above methods
int main()
{
string expression = "()())()";
removeInvalidParenthesis(expression);
expression = "()v)";
removeInvalidParenthesis(expression);
return 0;
} Java
// Java program to remove invalid parenthesis
import java.util.*;
class GFG
{
// method checks if character is parenthesis(open
// or closed)
static boolean isParenthesis(char c)
{
return ((c == '(') || (c == ')'));
}
// method returns true if string contains valid
// parenthesis
static boolean isValidString(String str)
{
int cnt = 0;
for (int i = 0; i < str.length(); i++)
{
if (str.charAt(i) == '(')
cnt++;
else if (str.charAt(i) == ')')
cnt--;
if (cnt < 0)
return false;
}
return (cnt == 0);
}
// method to remove invalid parenthesis
static void removeInvalidParenthesis(String str)
{
if (str.isEmpty())
return;
// visit set to ignore already visited string
HashSet visit = new HashSet();
// queue to maintain BFS
Queue q = new LinkedList<>();
String temp;
boolean level = false;
// pushing given string as
// starting node into queue
q.add(str);
visit.add(str);
while (!q.isEmpty())
{
str = q.peek(); q.remove();
if (isValidString(str))
{
System.out.println(str);
// If answer is found, make level true
// so that valid string of only that level
// are processed.
level = true;
}
if (level)
continue;
for (int i = 0; i < str.length(); i++)
{
if (!isParenthesis(str.charAt(i)))
continue;
// Removing parenthesis from str and
// pushing into queue,if not visited already
temp = str.substring(0, i) + str.substring(i + 1);
if (!visit.contains(temp))
{
q.add(temp);
visit.add(temp);
}
}
}
}
// Driver Code
public static void main(String[] args)
{
String expression = "()())()";
removeInvalidParenthesis(expression);
expression = "()v)";
removeInvalidParenthesis(expression);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to remove invalid parenthesis
# Method checks if character is parenthesis(open
# or closed)
def isParenthesis(c):
return ((c == '(') or (c == ')'))
# method returns true if contains valid
# parenthesis
def isValidString(str):
cnt = 0
for i in range(len(str)):
if (str[i] == '('):
cnt += 1
elif (str[i] == ')'):
cnt -= 1
if (cnt < 0):
return False
return (cnt == 0)
# method to remove invalid parenthesis
def removeInvalidParenthesis(str):
if (len(str) == 0):
return
# visit set to ignore already visited
visit = set()
# queue to maintain BFS
q = []
temp = 0
level = 0
# pushing given as starting node into queu
q.append(str)
visit.add(str)
while(len(q)):
str = q[0]
q.pop()
if (isValidString(str)):
print(str)
# If answer is found, make level true
# so that valid of only that level
# are processed.
level = True
if (level):
continue
for i in range(len(str)):
if (not isParenthesis(str[i])):
continue
# Removing parenthesis from str and
# pushing into queue,if not visited already
temp = str[0:i] + str[i + 1:]
if temp not in visit:
q.append(temp)
visit.add(temp)
# Driver Code
expression = "()())()"
removeInvalidParenthesis(expression)
expression = "()v)"
removeInvalidParenthesis(expression)
# This code is contributed by SHUBHAMSINGH10C#
// C# program to remove invalid parenthesis
using System;
using System.Collections.Generic;
class GFG
{
// method checks if character is
// parenthesis(open or closed)
static bool isParenthesis(char c)
{
return ((c == '(') || (c == ')'));
}
// method returns true if string contains
// valid parenthesis
static bool isValidString(String str)
{
int cnt = 0;
for (int i = 0; i < str.Length; i++)
{
if (str[i] == '(')
cnt++;
else if (str[i] == ')')
cnt--;
if (cnt < 0)
return false;
}
return (cnt == 0);
}
// method to remove invalid parenthesis
static void removeInvalidParenthesis(String str)
{
if (str == null || str == "")
return;
// visit set to ignore already visited string
HashSet visit = new HashSet();
// queue to maintain BFS
Queue q = new Queue();
String temp;
bool level = false;
// pushing given string as
// starting node into queue
q.Enqueue(str);
visit.Add(str);
while (q.Count != 0)
{
str = q.Peek(); q.Dequeue();
if (isValidString(str))
{
Console.WriteLine(str);
// If answer is found, make level true
// so that valid string of only that level
// are processed.
level = true;
}
if (level)
continue;
for (int i = 0; i < str.Length; i++)
{
if (!isParenthesis(str[i]))
continue;
// Removing parenthesis from str and
// pushing into queue,if not visited already
temp = str.Substring(0, i) +
str.Substring(i + 1);
if (!visit.Contains(temp))
{
q.Enqueue(temp);
visit.Add(temp);
}
}
}
}
// Driver Code
public static void Main(String[] args)
{
String expression = "()())()";
removeInvalidParenthesis(expression);
expression = "()v)";
removeInvalidParenthesis(expression);
}
}
// This code is contributed by Princi Singh 输出：
(())()
()()()
(v)
()v