对于y总解法的理解。

#include <cstdio>
#include <stack>
#include <cctype>
#include <unordered_map>
#include <cstring>
using namespace std;

const int maxn = 100010;
char expression[maxn];

void eval(stack<int>& num, stack<char>& op)
{
    int numaTmp = num.top(); num.pop();
    int numbTmp = num.top(); num.pop();
    char opTmp = op.top(); op.pop();

    int resTmp; //由于出栈顺序和入栈顺序相反，运算时要调换两数顺序
    if(opTmp == '+') resTmp = numbTmp+numaTmp;
    else if(opTmp == '-') resTmp = numbTmp-numaTmp;
    else if(opTmp == '*') resTmp = numbTmp*numaTmp;
    else resTmp = numbTmp/numaTmp;
    num.push(resTmp);
}

int main(void)
{
    unordered_map<char, int> priority{{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}};  //定义运算符的优先级
    stack<int> num; //存放运算式中的数字及中间结果
    stack<char> op; //存放运算式中的符号
    scanf("%s", expression);
    int len = strlen(expression);

    for(int i = 0; i < len; i++)
    {
        if(isdigit(expression[i]))
        {
            int x = 0;
            while(i < len && isdigit(expression[i]))
                x = x * 10 + expression[i++] - '0';
            num.push(x);    
            i--;    //遇到不是数字的字符需要回退
        }
        else if(expression[i] == '(') 
            op.push(expression[i]);
        else if(expression[i] == ')')
        {
            while(op.top() != '(') 
                eval(num, op);
            op.pop();   //括号内表达式计算完之后要让左括号出栈
        }
        else
        {
            //优先级高的运算符先运算并保证合法使用运算符
            while(op.size() && priority[op.top()] >= priority[expression[i]]) eval(num, op);    //这里的while不能用if，否则最后一个while中会死循环，而最后一个while又不可缺少，所以这里必须用while
            op.push(expression[i]);
        }  
    }
    while(op.size()) eval(num, op); //处理for循环中剩余的运算
    printf("%d\n", num.top());

    return 0;
}