栈操作的问题

题意：给出一个表达式，包含运算符+,-,*，min,max，求其结果

思路：用两个栈，一个操作数栈，一个操作符栈，当处理+,-,*操作符时，如果处理的操作符优先级比栈顶操作符优先级低，就入栈，否则，出栈；在处理min,max操作符时，如果当前处理的为min或者max,将min或者max,(两个入栈，当处理逗号时，就出栈，直到栈顶操作符为左括号，并将逗号入栈；而当处理右括号时，出栈直到栈顶为逗号。当处理完毕后，操作符栈可能不为空，需要出栈处理

代码如下：

#include <iostream>
#include <stack>
#include <string>
#include <map>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <fstream>

using namespace std;

const int MAX_LEN = 100;

class Solution
{
public:
int Calculate(char *pszExpr)
{
map<string, int> pPriority;
pPriority["+"] = 1;
pPriority["-"] = 1;
pPriority["*"] = 2;
pPriority["("] = 0;

stack<string> operStack;
stack<int> numStack;

int len = strlen(pszExpr);

for (int i = 0; i < len; i++)
{
if (isspace(pszExpr[i])) continue;

//数字
if (isdigit(pszExpr[i]))
{
int j = i;
int sum = 0;
while (j < len && isdigit(pszExpr[j]))
{
sum = sum * 10 + (pszExpr[j] - '0');
j++;
}
numStack.push(sum);
i = j - 1;
continue;
}

if (pszExpr[i] == '-' || pszExpr[i] == '+' || pszExpr[i] == '*')
{
if (operStack.empty())
{
string tmp;
tmp += pszExpr[i];
operStack.push(tmp);
}
else
{
string curOper;
curOper += pszExpr[i];
if (pPriority[curOper] > pPriority[operStack.top()])
{
operStack.push(curOper);
}
else
{
while(!operStack.empty() && pPriority[curOper] <= pPriority[operStack.top()])
{
int ans = popCal(numStack, operStack);
numStack.push(ans);
}
operStack.push(curOper);
}
}
}
else if (pszExpr[i] == 'm')
{
char tmp[4] = { 0 };
strncpy(tmp, &pszExpr[i], 3);
string strtmp = tmp;
operStack.push(strtmp);
operStack.push("(");
i += 3;
continue;
}
else if (pszExpr[i] == ',')
{
while (!operStack.empty() && operStack.top() != "(")
{
int ans = popCal(numStack, operStack);
numStack.push(ans);
}
operStack.push(",");
}
else if (pszExpr[i] == ')')
{
while (!operStack.empty() && operStack.top() != ",")
{
int ans = popCal(numStack, operStack);
numStack.push(ans);
}
operStack.pop();//弹出,
operStack.pop();//弹出(
int ans = popCal(numStack, operStack);
numStack.push(ans);
}
}

while (!operStack.empty())
{
int ans = popCal(numStack, operStack);
numStack.push(ans);
}

return numStack.top();
}

private:
int cal(int num1, int num2, string oper)
{
if (oper == "+")
{
return num1 + num2;
}
else if (oper == "-")
{
return num1 - num2;
}
else if (oper == "*")
{
return num1 * num2;
}
else if (oper == "min")
{
return min(num1, num2);
}
else if (oper == "max")
{
return max(num1, num2);
}
}

int popCal(stack<int>& numstack, stack<string>& operstack)
{
int num2 = numstack.top(); numstack.pop();
int num1 = numstack.top(); numstack.pop();
string op = operstack.top(); operstack.pop();
int ans = cal(num1, num2, op);

return ans;
}
};

验证代码如下：

1 * 2 + max(3,4)

min(2+3, max(4, 5))

int main()
{
#ifndef ONLINE_JUDGE
ifstream fin("f:\\oj\\uva_in.txt");
streambuf *old = cin.rdbuf(fin.rdbuf());
#endif
Solution solver;
char s[MAX_LEN];
while (cin.getline(s, MAX_LEN))
{
cout << "s:" << s << endl;
int ans = solver.Calculate(s);
cout << "ans:" << ans << endl;
}
#ifndef ONLINE_JUDGE
cin.rdbuf(old);
#endif
return 0;
}