Basic Calculator - LeetCode 224

题目描述：
Implement a basic calculator to evaluate a simple expression string.
The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and 
empty spaces .
You may assume that the given expression is always valid.
Some examples:
"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23
Note: Do not use the eval built-in library function.
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分析：
该题要实现一个带有括号优先级的加减法计算器，实质就是栈的使用。不多说了，直接看代码注释吧

/**///////////////////////56ms//*/class Solution {public:    int docompute(int a,int b,char opd){ //计算a opera bif(opd == '+')    return a + b;if(opd == '-')    return a - b;}    bool isopd(char ch){ //判断是否是运算符return (ch == '+' || ch == '-' || ch == '(' || ch == ')'||ch == '=');    }    int calculate(string s) {int len = s.size();if(len == 0)    return 0;s.push_back('='); //在表达式末尾加上'='，方便判断是否计算结束        stack<char> opd;opd.push('='); //在运算符栈底添加'='，方便判断是否计算结束stack<int> num; //操作数栈int i = 0;while( i <= len){     if(isdigit(s[i])){ // 如果是数字字符，则转换为整数后压入操作数栈。注意处理多位数的情况int dat = 0;        while(isdigit(s[i])){    dat = dat * 10 + (s[i]-'0');i++;}num.push(dat);    }    else if(isopd(s[i])){ //如果是运算符 if(s[i] == '('){  '('直接压入栈    opd.push(s[i]);    i++;    continue;}if(s[i] == ')'){                  //遇到')'时，需判断栈顶运算符是'+'或'-'，直接取出计算后将结果要入操作数栈；                //否则只能是'('，弹出栈.                //继续处理下一个字符    char oper = opd.top();     opd.pop();    if(oper == '+' || oper == '-'){                int a = 0, b = 0;b = num.top();num.pop();a = num.top();num.pop();int re = docompute(a,b,oper);num.push(re);opd.pop(); //pop teh '('    }    i++;     continue;}        if(s[i] == '+' || s[i] == '-'){  //遇到'+'或‘-’，看运算符栈顶    char oper = opd.top();     if(oper == '=' || oper == '('){ //若运算符栈顶是'='或'('，直接将当前运算符压入栈opd.push(s[i]);i++;continue;    }    else{  //栈顶为'+'或'-',取出计算opd.pop();int a = 0, b = 0;b = num.top();num.pop();a = num.top();num.pop();int re = docompute(a,b,oper);num.push(re);continue;    }}if(s[i]=='='){ //遇到'='，    if(opd.top() == '=') //栈顶也是'='，表示计算结束，返回结果return num.top();    else{ //不是等于，表明还有运算符，且只能是'+'或'-',取出计算结果char oper = opd.top();opd.pop();int a = 0, b = 0;b = num.top();        num.pop();a = num.top();num.pop();int re = docompute(a,b,oper);//cout << re << endl;num.push(re);continue;    }}            }     else //跳过空格i++;        }    }};