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📜  检查括号深度在给定的字符串中是否正确

📅  最后修改于: 2021-04-27 17:55:09             🧑  作者: Mango

给定由开括号,闭括号和整数组成的字符串S ,如果在此字符串中整数正确指示其深度,则任务是打印“是”。深度是指围绕该整数的一组嵌套括号的数量。否则打印否。

例子:

方法:

  1. 迭代由字符字符串的字符
  2. 如果要打开或关闭括号,则将其追加到数组arr []中
  3. 如果字符是数字,则迭代直到读取整个数字,然后追加到arr []中
  4. 逐元素迭代数组arr []元素
    • 如果元素为’(’,则增加深度并以1打开
    • 如果元素为’)’,则将深度减1,然后将深度减1
    • 如果元素是整数,则检查“ Integer!= depth”,如果为true
      将标志设置为0并打破循环
  5. 遍历整个字符串,检查open是否等于close,如果true,则设置标志= 0

下面是上述方法的实现

C++
// Function to check if the Depth
// of Parentheses is correct
// in the given String
  
#include 
using namespace std;
  
bool Formatted(string s)
{
    vector k;
    int i = 0;
  
    while (i < s.size())
    {
        // Appending if the
        // Character is not integer
        if (s[i]==')' or s[i]=='(')
        {
            k.push_back(s[i]);
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            char st;
            while (s[i]!=')' and s[i]!=')')
            {
                st = s[i];
                i = i + 1;
            }
            k.push_back(st);
        }
    }
  
    int depth = 0, flag = 1;
    int open = 0, close = 0;
  
    for (char i:k)
    {
  
        // Check if character is '('
        if (i == '(')
        {
            // Increment depth by 1
            depth += 1;
  
            // Increment open by 1
            open += 1;
  
        // Check if character is ')'
    }else if (i == ')'){
  
            // Decrement depth by 1
            depth -= 1;
  
            // Increment close by 1
            close += 1;
        }
        else{
            if (i-'0' != depth){
                flag = 0;
                break;
            }
        }
    }
  
    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;
  
    return (flag == 1)?true:false;
}
  
// Driver Code
int main()
{
    string s = "((2)((3)))";
    bool k = Formatted(s);
    if (k == true)
        printf("Yes");
    else
        printf("No");
    return 0;
}
  
// This code is contributed by mohit kumar 29


Java
// Function to check if the Depth
// of Parentheses is correct
// in the given String
import java.util.*;
  
class GFG{
  
static boolean Formatted(String s)
{
    Vector k = new Vector();
    int i = 0;
  
    while (i < s.length())
    {
        // Appending if the
        // Character is not integer
        if (s.charAt(i)==')' || s.charAt(i)=='(')
        {
            k.add(s.charAt(i));
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            char st = 0;
            while (s.charAt(i)!=')' && s.charAt(i)!=')')
            {
                st = s.charAt(i);
                i = i + 1;
            }
            k.add(st);
        }
    }
  
    int depth = 0, flag = 1;
    int open = 0, close = 0;
  
    for (char i2 : k)
    {
  
        // Check if character is '('
        if (i2 == '(')
        {
            // Increment depth by 1
            depth += 1;
  
            // Increment open by 1
            open += 1;
  
        // Check if character is ')'
    }else if (i2 == ')'){
  
            // Decrement depth by 1
            depth -= 1;
  
            // Increment close by 1
            close += 1;
        }
        else{
            if (i2-'0' != depth){
                flag = 0;
                break;
            }
        }
    }
  
    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;
  
    return (flag == 1)?true:false;
}
  
// Driver Code
public static void main(String[] args)
{
    String s = "((2)((3)))";
    boolean k = Formatted(s);
    if (k == true)
        System.out.printf("Yes");
    else
        System.out.printf("No");
}
}
  
// This code is contributed by Rajput-Ji


Python3
# Function to check if the Depth 
# of Parentheses is correct
# in the given String
  
def Formatted(s):
    k = []
    i = 0
  
    while i < len(s):
  
        # Appending if the 
        # Character is not integer
        if s[i].isdigit() == False:
  
            k.append(s[i])
            i += 1
        else:
  
            # Iterating till the entire
            # Digit is read
            st = ""
            while s[i].isdigit():
                st += s[i]
                i = i + 1
            k.append(int(st))
  
    depth, flag = 0, 1
    open, close = 0, 0
  
    for i in k:
  
        # Check if character is '('
        if i == '(':
  
            # Increment depth by 1
            depth += 1
  
            # Increment open by 1
            open += 1
  
        # Check if character is ')'
        elif i == ')':
  
            # Decrement depth by 1
            depth -= 1
  
            # Increment close by 1
            close += 1
        else:
            if i != depth:
                flag = 0
                break
  
    # Check if open parentheses
    # NOT equals close parentheses
    if open != close:
        flag = 0
  
    return True if flag == 1 else False
  
# Driver Code
if __name__ == '__main__':
    s = '((2)((3)))'
    k = Formatted(s)
    if k == True:
        print("Yes")
    else:
        print("No")


C#
// Function to check if the Depth
// of Parentheses is correct
// in the given String
using System;
using System.Collections.Generic;
  
class GFG
{
  
static bool Formatted(String s)
{
    List k = new List();
    int i = 0;
  
    while (i < s.Length)
    {
        // Appending if the
        // char is not integer
        if (s[i]==')' || s[i]=='(')
        {
            k.Add(s[i]);
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            char st = '\x0000';
            while (s[i] != ')' && s[i] != ')')
            {
                st = s[i];
                i = i + 1;
            }
            k.Add(st);
        }
    }
  
    int depth = 0, flag = 1;
    int open = 0, close = 0;
  
    foreach (char i2 in k)
    {
  
        // Check if character is '('
        if (i2 == '(')
        {
            // Increment depth by 1
            depth += 1;
  
            // Increment open by 1
            open += 1;
  
        // Check if character is ')'
    }else if (i2 == ')'){
  
            // Decrement depth by 1
            depth -= 1;
  
            // Increment close by 1
            close += 1;
        }
        else{
            if (i2-'0' != depth){
                flag = 0;
                break;
            }
        }
    }
  
    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;
  
    return (flag == 1)?true:false;
}
  
// Driver Code
public static void Main(String[] args)
{
    String s = "((2)((3)))";
    bool k = Formatted(s);
    if (k == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by 29AjayKumar


输出:
Yes