我正在嘗試將中綴表達式轉換爲後綴表達式的代碼。目前,如果我輸入例如「5 + 6」,程序將正常工作,它將輸出正確的答案,即「5 6 +」。當我輸入多於一個運算符(例如「5 + 6-3」)時,會出現問題,輸出錯誤答案「+ 3-」。有人可以指出我犯的錯誤嗎?提前致謝 !中綴到Postfix轉換
void main(){
Stack *s = new Stack;
string input;
cout <<"Enter Expression"<<endl;
cin>>input;
InfixToPostfix(input);
system("PAUSE");
}
string InfixToPostfix(string input){
Stack *S = new Stack();
string postfix = "";
for (int i=0; i < input.length();i++){
if (input[i]== ' '||input[i]==',') continue;
else if (IsOperator(input[i]))
{
while(!S->IsStackEmpty() && S->StackTop() != '(' && HasHigherPrecedence(S->StackTop(),input[i]))
{
postfix=S->StackTop();
S->Pop();
}
S->Push(input[i]);
}
else if(IsOperand(input[i]))
{
postfix +=input[i];
}
else if (input[i] == '(')
{
S->Push(input[i]);
}
else if (input[i]==')')
{
while(!S->IsStackEmpty() && S->StackTop() != '('){
postfix += S->StackTop();
S->Pop();
}
S->Pop();
}
}
while(!S->IsStackEmpty()){
postfix +=S->StackTop();
S->Pop();
}
cout <<""<<postfix;
return postfix;
}
bool IsOperand(char C)
{
if(C>= '0' && C<= '9') return true;
if(C>= 'a' && C<= 'z') return true;
if(C>= 'A' && C<= 'Z') return true;
return false;
}
bool IsOperator(char C)
{
if(C=='+' || C== '-' || C =='*' || C == '/' ||C == '$')
{
return true;
}else{
return false;
}
}
int IsRightAssociative(char op)
{
if(op=='$'){
return true;
}else{
return false;
}
}
int GetOperatorWeight(char op){
int weight = -1;
switch(op)
{
case'+':
case '-':
weight=1;
break;
case '*':
case '/':
weight=2;
break;
case '$':
weight=3;
break;
}
return weight;
}
int HasHigherPrecedence (char op1, char op2)
{
int op1Weight= GetOperatorWeight(op1);
int op2Weight = GetOperatorWeight(op2);
if(op1Weight == op2Weight)
{
if(IsRightAssociative(op1))
{
return false;
}else{
return true;
}
return op1Weight > op2Weight ? true:false;
}
}
這看起來不像C++/CLI。它看起來像普通的C++和一個未知的Stack類... – Medinoc