栈操作的问题

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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_JUDGEifstream fin("f:\\oj\\uva_in.txt");streambuf *old = cin.rdbuf(fin.rdbuf());#endifSolution 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_JUDGEcin.rdbuf(old);#endifreturn 0;}

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