任何人都可以帮助我使这段代码工作 [英] Can anyone help me to make this code to work

问题描述

我正在尝试平衡这些字符，我收到一些错误，任何人都可以尝试查看我需要更改的代码



错误是分段错误： 11



我尝试过：

i am trying to balance these characters , i receiving some errors , can anyone try to see my code where i need the changes

Error is "Segmentation fault: 11"

What I have tried:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef char T;
#include "stack.h"

int isBalanced(char s[100])
{
    Stack *stack;
    int n=strlen(s);
    int i;
    for(i=0;i<n;i++)
    {
      if(s[i]=='(' || s[i]=='[' || s[i]=='{')
      {
        Push(&stack,s[i]);
      }
      else if(s[i]==')' || s[i]==']' || s[i]=='}')
      {
        if(IsEmptyStack(stack))
        {
          return 0;
        }
        else if(Top(stack)!=s[i])
          Pop(stack);     
      }
    }
    if(IsEmptyStack(stack))
      return 1;
    else
      return 0;
   

}

int main()
{
    char s1[100]="((()))", s2[100]="((())(";

    if(isBalanced(s1))
      printf("Balanced | \n");
    else
      printf("Not balanced | \n");

    if(isBalanced(s2))
      printf("Balanced | \n");
    else
      printf("Not balanced | \n");

    return 0;
}

推荐答案

你的意思是

Do you mean

#include <iostream>
#include <cstring>
#include <stack>
using namespace std;

int isBalanced(char s[100])
{
    stack<char> stk;
    int n=strlen(s);
    int i;
    for(i=0;i<n;i++)
    {
      if(s[i]=='(' || s[i]=='[' || s[i]=='{')
      {
        stk.push(s[i]);
      }
      else if(s[i]==')' || s[i]==']' || s[i]=='}')
      {
        if(stk.empty())
        {
          return 0;
        }
        else if(stk.top() != s[i])//<--- What's the purpose of this line?
          stk.pop();
      }
    }
    if(stk.empty())
      return 1;
    else
      return 0;
}

然而我无法猜出标记线的用途

However I can't guess the purpose of the marked line

你的算法错了。

当你遇到一个开头（[{，你把它推到堆栈上，到目前为止，那么好。

当你遇到收盘）]} 时，你检查堆栈是否为空，好吧。

然后你应该检查一下右括号与堆栈上的开头匹配，否则失败。

如果（位于堆栈顶部，）是唯一允许的。



您必须重新考虑您的算法。

遇到时a，打开括号，我会在堆栈上推送结束副本，它会简化以后的检查。

Your algorithm is wrong.
When you encounter an opening ([{, you push it on stack, so far, so good.
When you encounter a closing )]}, you check if stack is empty, ok.
then you should check if the closing parenthesis is matching the opening one on stack, otherwise, it is a fail.
if ( is on top of stack, ) is the only one allowed.

You have to rethink your algorithm.
When encountering a, opening parenthesis, I would push the closing counterpart on stack, it simplify later checking.

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