我上周收到了 C++ 类(class)的作业。我想你们中的一些人会觉得它很有趣!我设法写下了大部分代码，但我被卡住了，无法在我的一生中解决这个问题……以下是我必须放入代码中的加密过程的指南:

The message sender inputs a four letter word, CCCC, and another four letter word, XXXX.

The message sender then inputs the message to be encrypted.

The program scans the message one char at a time and each char is pushed in a stack until either the scanned character is in the word CCCC or the end of the message is encountered.

When the scanned character is one of the chars in CCCC, print that char and continue to print and pop the chars at the top of the stack until either the stack is empty or the char at the top of the stack is one of the chars in XXXX. When the end of the message is encountered, print the character at the top of the stack and continue to pop and print from the top of the stack until the stack is empty.

Here is a hint: "GOOD" "LUCK", it "SOUNDS SIMPLE TO ME", or as your program would say: "OSDNOT EEM LPMIS SU"

这就是实际的任务。

我遇到的问题是最后一点:

When the end of the message is encountered, print the character at the top of the stack and continue to pop and print from the top of the stack until the stack is empty.

现在这是我到目前为止的代码:

#include <string>
#include <iostream>
using namespace std;
class Stack
{
   private:
   char Chars[50];
   int top;
   public:
   int push(char);
   char pop();
   bool isEmpty();
   bool isFull();
   Stack()
   {
      top = 0;
   }
};

int main()
{
Stack theStack;
   char word1[4];
   char word2[4];
   for(int i=0; i < 4; i++){
      word1[i] = ' ';
      word2[i] = ' ';
   }
   char message[500];   
   cout << "Please enter a 4 letter word: ";
   cin >> word1;
   while(word1[4] || !word1[3])
   {
      cout << "Word must be 4 chars long. Try again: ";
      cin >> word1;
   }
   cout << "Please enter another 4 letter word: ";
   cin >> word2;
   while(word2[4] || !word2[3])
   {
      cout << "Word must be 4 chars long. Try again: ";
      cin >> word2;
   }
   cout << "Please enter the phrase to be encrypted (50 chars max): ";
   cin.ignore(1000, '\n');
   cin.getline(message,500);
   int length = strlen(message);
   int count = 0;
   char finalMsg[length];
   //scanner
   for(int i = 0; i < length; i++)
   {
      if(message[i] == word1[0] ||
         message[i] == word1[1] ||
         message[i] == word1[2] ||
         message[i] == word1[3])
      {
         finalMsg[count] = message[i];
         count++;
         if(message[i-1] != word2[0] ||
            message[i-1] != word2[1] ||
            message[i-1] != word2[2] ||
            message[i-1] != word2[3])
         {
            finalMsg[count] =  message[i-1];
            count++;
         }
      }
      else
      {
         theStack.push(message[i]);
      }
   }
   cout << finalMsg << endl;
return 0;
}

int Stack::push(char data)
{
   Chars[top] = data;
   top++;
return top;
}

char Stack::pop()
{
   char ret = Chars[top-1];
   top--;
return ret;
}

bool Stack::isEmpty()
{
   if(top <= 0)
      return true;
   else return false;
}

bool Stack::isFull()
{
   if(top >= 50)
      return true;
   else return false;
}

编译后，最终输出为我提供了“OSDNOT”，这是我教授提供的示例中的内容，所以我知道我正朝着正确的方向前进。任何帮助都会很棒，我什至不知道从哪里开始检查代码。

最佳答案

这是更正后的代码。您没有正确编写算法代码。我已经评论了我在代码中所做的更改。 首先，当你在扫描时遇到CCCC中存在的字符时，你没有弹出堆栈的元素。同样在扫描结束时，您没有清空堆栈。包括 cstring 而不是 string。正如评论中所指出的，您对 word1 和 word2 的声明不正确。

   char finalMsg[200];
   //scanner
   for(int i = 0; i < length; i++)
   {
      if(message[i] == word1[0] ||
         message[i] == word1[1] ||
         message[i] == word1[2] ||
         message[i] == word1[3])
      {
         finalMsg[count] = message[i];
         count++;

         //pop out elements from the stack till it is empty or an character of XXXX is encountered
         while(!theStack.isEmpty())     
         {
             char tmp=theStack.pop();
             if(tmp==word2[0] ||
                tmp==word2[1] ||
                tmp==word2[2] ||
                tmp==word2[3])
                {
                    theStack.push(tmp);
                    break;
                }
            finalMsg[count++]=tmp;
         }

      }
      else
      {
         theStack.push(message[i]);
      }
   }

  //empty the stack
   while(!theStack.isEmpty())
   {
       finalMsg[count++]=theStack.pop();
   }
   finalMsg[count++]=0;
   cout << finalMsg << endl;

PS:最好使用 std::stack 和 std::string。

关于c++ - 使用堆栈加密字符串，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11007719/