山东大学数据结构实验 栈——计算表达式

建立两个栈
数字栈和符号栈。
读入表达式
遇到数字存入数字栈,遇到左括号和±*/存入符号栈
维护符号栈中符号优先度单调。
如果遇到右括号,那么一直向右计算,直到遇到左括号为止
代码:

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int level[200];//符号优先级
template<class T>
class Stack{
    
    
	public:
		Stack(){
    
    
			change_len(10);			
		}
		void change_len(int change_size)//更新长度 
		{
    
    			
			T * tmp = new T[change_size];//新建一个新动态数组 
			int num1 = min(_size,change_size);
			copy(sk,sk+num1,tmp);//复制到新数组 			
			sk = tmp;
			_size = change_size;//更新长度 
		}
		int size(){
    
    
			return this->cap;
		}
		bool empty(){
    
    
			return (cap==0);
		}
		void push(T value){
    
    
			sk[cap++] = value;
			if(cap==_size){
    
    
				change_len(_size*2);
			}		
			stacktop = value;	
		}
		void pop(){
    
    
			if(this->size()==0)return ;
			else {
    
    
				cap--;
				sk[cap] = 0;
				if(cap!=0)stacktop = sk[cap-1];
				else stacktop = 0;
			}
		}
		T top(){
    
    
			return stacktop;
		}
	private:
		T* sk;//栈 
		T stacktop;//栈顶 
		int _size=0;//空间 
		int cap=0;//容量 
};
 

//计算 
void calculate(Stack<double>&num,Stack<char>&opera){
    
    
    double n1 = num.top();//取数字栈中的两个数
    num.pop();
    double n2 = num.top();//取数字栈中的两个数
    num.pop();
    double ans;
    switch (opera.top()) {
    
    
        case '+':ans = n2+n1;  break;
        case '-':ans = n2-n1;  break;
        case '*':ans = n2*n1; break;
        case '/':ans = n2/n1;
    }
    num.push(ans);//将新数返回数字栈
    opera.pop();//符号栈弹栈
}


void pushStack(string s,Stack<double>&num,Stack<char>&opera){
    
    
    int len = (int)s.length();
	double sum = 0;
    for(int i = 0;i < len;i++){
    
    
        if(isdigit(s[i])){
    
    //处理数字
            sum = sum*10.0+(double)(s[i]-'0');
            if(i == len-1||!isdigit(s[i+1]) ) 
			{
    
    
				num.push((double)sum);
				sum = 0;
			}
        }
        //处理运算符
        else {
    
    
            if(s[i] == '(') {
    
    //左括号直接入栈
                opera.push('(');
                continue;
            }else if(s[i] == ')'){
    
    //右括号要不断弹出符号栈,直到遇到左括号
                while(!opera.empty() && opera.top() != '(') calculate(num,opera);
                continue;
            }
            //计算栈顶的情况 
            while(!opera.empty() && level[s[i]] <= level[opera.top()])  calculate(num,opera);
            opera.push(s[i]);
        }
    }
}

double getResult(string &s,Stack<double>&num,Stack<char>&opera){
    
    
    pushStack(s,num,opera);
    while(!opera.empty()){
    
    
        calculate(num,opera);
    }
    return num.top();
}

int main(){
    
    
	
	string s="";//需要运算的字符串 
	
	Stack<double>num;//数字栈
	Stack<char>opera;//符号栈

    level['/'] = 3;//运算符优先级
	level['*'] = 3;
	level['+'] = 2;
	level['-'] = 2;
    int n;
	double ans;
    cin>>n;
    for(int i = 0;i < n;i++){
    
    
        while(!num.empty()) num.pop();//清空栈
        while(!opera.empty()) opera.pop();
        cin>>s;
		printf("%.2lf\n",getResult(s,num,opera) );
    }
    return 0;
}

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转载自blog.csdn.net/naiue/article/details/109320334
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