题解 | #计算表达式#化简了点代码

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <stack>
#include <map>
using namespace std;

map<char, int> mymap = {
	{'+', 0},{'-', 0},{'/', 1},{'*', 1}
};

float jisusan(float x, float y, char oper){
	float res;
	switch(oper){
		case '+':
			res = x + y;
			break;
		case '-':
			res = x - y;
			break;
		case '*':
			res = x * y;
			break;
		case '/':
			res = x / y;
			break;
	}
	return res;
}


int main() {
    stack<float> numStack;
    stack<char> operStack; //operator 运算符 
    string expr; //expression 表达式 
    while(getline(cin, expr)){
    	for(int i = 0; i < expr.size(); i++){
    		if(expr[i] >= '0' && expr[i] <= '9'){ 
				//表明这是一个数字 可能有很多位 
				int j = i;
    			while(j < expr.size() && expr[j] >= '0' && expr[j] <= '9'){
    				j++;
    				continue;
				}
				//此时j越界，或者j是操作符 i 到 j - 1为数字
				float num = 0; 
				while(i != j){
					num *= 10;
					num += (expr[i] - '0');
					i++;
				}
				i--;
				numStack.push(num);
			}else{
				//运算符先跟栈里的比较
				//此次的运算符是一定要放进去的 
				while(!operStack.empty() && mymap[operStack.top()] >= mymap[expr[i]]){
				 	//栈顶的优先级 >= 当前的优先级 
					//弹出两个数和操作符计算结果然后放到numstack
					float a1 =  numStack.top();
					numStack.pop();
					float a2 = numStack.top();
					numStack.pop();
					char c = operStack.top();
					operStack.pop();
					float res = jisusan(a2, a1, c);
					numStack.push(res);	
				}

				operStack.push(expr[i]);				
			}	
		}
		//把剩余的都弄出来 
		while(!operStack.empty()){
			float a1 =  numStack.top();
			numStack.pop();
			float a2 = numStack.top();
			numStack.pop();
			char c = operStack.top();
			operStack.pop();
			float res = jisusan(a2, a1, c);
			numStack.push(res);
		}
		float a1 =  numStack.top();
		numStack.pop();
		//printf("%f\n", a1); //这里用cout方便点 
		cout << a1 <<endl;
	
    }
}
// 64 位输出请用 printf("%lld")