Question

我被告知编写一个程序，使用堆栈将前缀形式转换为postfix形式当我使用纸和笔来实现该功能时，我现在的输出应该是正确的。但是，命令窗口中显示的结果很奇怪。

实际输出：

prefix  : A
postfix : A

prefix  : +*AB/CD
postfix : AB*CD/+

prefix  : +-*$ABCD//EF+GH
postfix : AB$C*D-EF/GH+/H

prefix  : +D/E+$*ABCF
postfix : DEAB*C$F+/F

prefix  : /-*A+BCD-E+FG
postfix : ABC+DEFG+-+FG

正确输出：

prefix  : A
postfix : A

prefix  : +*AB/CD
postfix : AB*CD/+

prefix  : +-*$ABCD//EF+GH
postfix : AB$C*D-EF/GH+/+

prefix  : +D/E+$*ABCF
postfix : DEAB*C$F+/+

prefix  : /-*A+BCD-E+FG
postfix : ABC+*D-EFG+-/

代码：

void prefix_to_postfix(string& prefix, string& postfix)
{
//Convert the input prefix expression to postfix format

postfix = prefix;   //initialize the postfix string to the same length of the         prefix string

stack<stackItem> S;
stackItem x;
int k = 0;  //index for accessing char of the postfix string

for (int i = 0; i < prefix.length(); i++)  //process each char in the prefix string from left to right
{
    char c = prefix[i];

    if(prefix.length() == 1)
        break;

    //Implement the body of the for-loop        
    if(isOperator(c))
    {
        x.symb = c;
        x.count = 0;
        S.push(x);
    }
    else
    {
        S.top().count++;
        postfix[k++] = c;

        if(S.top().count == 2)
        {
            postfix[k++] = S.top().symb;
            S.pop();
            S.top().count++;
        }
    }
    if(i == (prefix.length() - 1))
    {
        postfix[k++] = S.top().symb;
        S.pop();
    }

}
}

Answer 1

您似乎熟悉OOP基础知识，因此我建议采取更清洁的方法。对我来说，首先从前缀生成一个树然后通过左右会第一次迭代来获得后缀似乎更好。

困难的部分是生成树，首先考虑使用名为TNode的结构：

class TNode
{
private:
    TNode* _left;
    TNode* _right;
public:
    TNode* Parent;
    char Symbol;
    bool IsOperand;
    TNode(char symbol , bool isOperand)
    {
        Symbol = symbol;
        IsOperand = isOperand;
        Parent = NULL;
        _left = NULL;
        _right = NULL;
    }     
    void SetRight(TNode* node)
    {
        _right = node;
        node->Parent = this;
    }

    void SetLeft(TNode* node)
    {
        _left = node;
        node->Parent = this;
    }

    TNode* GetLeft()
    {
        return _left;
    }

    TNode* GetRight()
    {
        return _right;
    }
};

这是树生成器：

TNode* PostfixToTree(string prefix)
{
    TNode* root = NULL;
    TNode* nodeIter = NULL;
    char c;
    for(int i=0 ; i<prefix.length() ; i++)
    {
        c = prefix[i];
        if(root == NULL)
        {
            if(!isOperand(c))
            {
                root = new TNode(c,false);
                nodeIter = root;
            }
            else return NULL;
        }
        else
        {
            while(true)
            {
                if(nodeIter->GetLeft() == NULL && !isOperand(nodeIter->Symbol))
                {
                    nodeIter->SetLeft(new TNode(c,isOperand(c)));
                    nodeIter = nodeIter->GetLeft();
                    break;
                }
                else if(nodeIter->GetRight() == NULL && !isOperand(nodeIter->Symbol))
                {
                    nodeIter->SetRight(new TNode(c,isOperand(c)));
                    nodeIter = nodeIter->GetRight();
                    break;
                }
                else
                {
                    while(isOperand(nodeIter->Symbol) ||
                        nodeIter->GetRight()!=NULL && nodeIter->GetLeft()!=NULL &&
                        nodeIter->Parent!=NULL)
                    {
                        nodeIter = nodeIter->Parent;
                    }
                }
            }

        }

    }

    return root;
}

最后是从树生成Postfix的函数。

string TreeToPostfix(TNode* root)
{
    string postfix = "";
    stack<TNode*> nodeStack;
    nodeStack.push(root);
    while(!nodeStack.empty())
    {
        TNode* top = nodeStack.top();
        nodeStack.pop();
        postfix = top->Symbol + postfix;
        if(top->GetLeft()!=NULL)
            nodeStack.push(top->GetLeft());
        if(top->GetRight()!=NULL)
            nodeStack.push(top->GetRight());
    }
    return postfix;
}

使用堆栈将前缀转换为postfix

1 个答案: