题目：
https://leetcode.com/problems/basic-calculator/
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.
思路一：
按照运算符优先级将原表达式转化为波兰表达式后缀式，然后再计算后缀式，与之前作业方法一致。
然而“前人”的博客已经写得很好了（见文末链接），上节课又刚好讲了递归，于是有了另一个作死的思路。
思路二：
计算只有加减的表达式，遇到括号递归调用该函数。关键部分如下：
for (i = 0; i < ori.size() - c; i++) { a = ori[i+c]; switch (a) { //为简洁删掉其余case和default部分的代码 case '(': str.assign(ori, i+c+1, ori.size()-i-c-1); c += getOp(str, m)+1; elem.e.m = m; elem.isChar = 0; op.push_back(elem); break; case ')': while (!s.empty()) { t = s.top(); s.pop(); elem.e.c = t; elem.isChar = 1; op.push_back(elem); } n = calOp(op); return i; break; } }
然而在括号过多时，算法复杂度为O(n2)，遇到文章最后的给出的输入会超时（此处不想吐槽……）
思路三：
对该题目，其实只需要判断每个数应该进行的操作('+'/'-')，于是可以只用一个用于记录当前（括号外）的加或减的栈，再对括号内每个数判断其前面的符号进行相应操作。这样只需遍历一次，复杂度O(n)。不过还要注意空格的问题。代码如下：
int calculate(string s) { stack<bool> isPos;//当前括号外应执行的操作，'+'为1 char c; bool curPos = 1;//当前待执行的操作，'+'为1 isPos.push(1);//初始操作为'+' int sum = 0, t; for (int i = 0; i < s.size(); i++) { c = s[i]; switch (c) { //判断当前应执行的操作 case '+': if (isPos.top() == 0) { curPos = 0; } else { curPos = 1; } break; case '-': if (isPos.top() == 0) { curPos = 1; } else { curPos = 0; } break; //遇到左括号更新括号外的操作 case '(': isPos.push(curPos); break; //遇到右括号弹出当前括号外的操作，恢复为上一个括号内 case ')': isPos.pop(); break; //遇到数字继续扫描至符号 case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9': t = 0; while (s[i] >= '0' && s[i] <= '9') { t = 10*t + s[i] - '0'; i++; } i--; if (curPos == 0) { sum -= t; } else { sum += t; } break; //忽略其他的符号（其实只有空格） default: break; } } return sum; }
”前人“们的博客：
http://200404.sinaapp.com/2015/10/evaluating-an-arithmetic-expression/http://www.cnblogs.com/lustralisk/p/brance-2.htmhttp://jmq14.github.io/2015/10/23/calculator/思路二没通过的例子：
"(9-(10-(10-0-(3+(8+(0+(8-(10-8-(7-(2+(5+(6+(10+(3+(8+(3-(9+(1+(10+(1-(1+(6-2+0+(10-(9-(3-(3-9-(1-(7+(4-(2+(2-(10+(3+(7-(1-(4+(1+(1-(10-(5-(9+(9-4-(5-(1+8-(2-(1+(1-10-(4-(1+(4-(7)-(3-(8)+(5+5-(5-(9-(8+(8-4-1+(0-(1+(1+(10-(7+(2-(5-(4-(6+(2+(1-(2-(9+8+(2+(9-(9-(7+(10+1+(5)))-(2-(8+3+(5-(7-(3+(9)+(10+(0+(8-(1-(9)-(0+10-(3+(9-(0-(5-(7-(4-4+1+(7)-(10+(5+(9-(3+(5+(6-(0-(7-(1-(4+(6+(4-2-(4+(9-(6+9-8+1+(5+(7-9+3)+(10-(10+(2+(0-(5-(2+(10-(4-5-(7-(4-(7+(4)+6+10+(2-(7+(2))+(1)+(5-(7)-(10-(5+(7-(6-(2+(1-4)+(10-(5)+(4+(10+(4+(0+(10+(8-(8+(6+5-(1-(6-(1-(2+(4+(9-(3+(1+(10+(4)+(0+(3-(2-(9-(2-(3-(4-(2+(7-(6-(5+(7+(5+(5-(4+(0-(7+(2-(7+(9)-(6-(10)+(7+(2-(9-(9)+(4+(1-(8+(2-0-(2+(2+10)-(7-9-(9+(8-(5-8-(5)+(6+(10-(3-(2-(2+(7-2+(9+(3+(9+(2-(8+(5-(4+(4-(1-(9+(0+(6-(4-(3+(5-(2-(4-(6+(0+(4+3)-(8-(6+(9+(1+(2)-(8-(1+1+(5+(4-(3-(1-(7-4+(6+(9+(1+(4)+(6+(4+(2+(7-(1+(4-(8+(6+(8-(9-(2)-3-(0-(0)+(5+(7-(8)+(8-(2+(1)+1+(3+(6-(10-(2-4-(2-(2)+(8)+(3-(1-(1)+(6+(1+(9+(9+(5)-(4+(9+(10)+(0-(3+(3+0)+(6)-(6+(6)+(4-(8-1-5-(6)-(0))-(3)+(3-(3-(8-(10-(0-(4+(7)+(6-4))+1-(2-(1-(0-(0+(1-(0)-0+(5+(10-(2-(9-(9-10)+(3+(5-(6-(6-9-(5+5))+(7+(0)-(2-(7+2+(7-(2+(7+(4-(10+(4+(10-(3-(0-2+(9+(4-4-(3-(2)+(8+(5)+(1+1-(7+(3+(10+5-(0+(10-(9+(8-(0-(0+(8-(1+(0)+(6+(5+(5+(9)))+(4-(1-(3+(7+(9+(8-(1-8-(8+(0+(1+(1-(1)+(7+(6-(7-(8+(10)+1+(0-(10)+(8+(7+(10+(6+(10+(6)-(2+(2+(10-(8)-(5)))+(9-(1)+(4)+(5)-(6-(9)-(1+(6-(9+(10)+2-(4+(9-(4+1)-(0-(9)-(3)+(0)+(10)))+9)+(6+4+(6))+(5-(9))-(9-(2-(6+(7))-(6-(3+(5+(5-(0)-(5+(6-(5+(9-(2+(9+(1+(0+2+(7)-(3-(5+(2)+(4)+(6+(7-(3-(4)+(10+(4))+(3))-(3-(2)-(2+(2+(10+(3)+(3+(5)-(3-(0+(1)+(6+(4-(4)-(7-(9-(9)+(1)+(4)+(7))-(9))))-(3-(1+5-7-(7))-(4+(3+(7-(9+(8)-(9+(8-(3)+(10-(1)+(5)-(2-(4)+(0-(10-(7-(10+(1)+(1)-(4)-(10)))+(7)+(4-4)+0+9-(6))-(6+(5)))))-(8-(6)-(10+(5-(8)-(10+(3+(0+(6-(9)-(1)))-(0)-(9+(0+(1+(8+2-(4-(9-(4+(3+4)-(10+(1-(5)+(10-(4-(6-(4-(2+(4)-(9)-(4))))-3))))+(9)+(9+(0-(1+(5-(5+(7)-6-(8-(3-(3+(1)-(9-(7-(6)))-(2+(1))-(1+(2+(10))))+(6)+(0+(9-(1)-(10)))+(10-(1-(1)))-(0+(0-(2-(4-(6+(1))+(0)+(5)-(5+(5)-(4+(6)-(5)+(1-(7))))+(8)-(7))-3))-(7+(7+(9+(0+(10)-(7-(0-(2)-(6))-(2+(10)))+(7)+(3))+(8-(8+(10)-(8)+(0+(6-(2)-(1))+(3+(10+(10-(4+(7-(2)-(9-(2+(8))))+(7)-(7+10+(9-(2)+(0))-(6+(1)))+(10)+(2)-(7)-(4)-(10+(3-(6))))+(8-(1))-(10)))+(5+(3-(0-(1-(2+(3-(6-(4)-(1)+(4+(7+(3)-(7)+(4-(9))+(0-(4)+(9+(3-(9)+(4-(10+(6+(4)))+(4))+(10+(0-3-(8+(0-(6))-(5))-(9))-(6))))+2))+6+(6)+(1-(6))-(7-(1))-(8)+(9-(8))+(4)))-(0+7-(1)))-(2))+(0)))+(4-(7))-(5)-(8)-4+(1-(3-(8+(2+0)+(7)))))))-(4-(2))+(9))))+(7)-(2-(10+(4)-(8+(7)+(5-(4)-(6+6))-(2+(6)-(2+(4-(2-(8-(4)-(7+(5)-(10-(7)))))-(10+(9+(8)-(10)+(3-(7+(4+(2+(5)-(10+7+(2-(10)-(10+(3))+(0-(10+(8+(4+(7-(2)+(3+9))))+(7-(6+(2)-(2)+7+(5+(7+(10+(5-(4)-2+(5)+(1))+(0))))-(9))-5-(8)-(9-(4)-(10))-(8-(5)-(10)-7)+(5))-(4)))))+6+3+(3+(6+(9)))-10+(6)+(0)))))+(7)))+(1-(5)+(3-(3+6))+(5)+(7)-(9-(1))+(4+(1))+(2)))-(3))-(10)+(1)))))))))+(3)+2+(8-(4)))-(1))+(6-(8-(0)-(8-(0))-(2-(4+2)))-(9+1)))-(8-(8+(1-8-(7))))+7-(5+(5+(6+(10)+(8)))))))-(4))))-(4)-(6)+(10)-(5)))+(0+(2+(4))-(4-(2)+(0-(10-(4))))))+3-(10)))-(9+(9-(8-(7)-4))))+(6))-(4-(9))))-(1))+(10))))-(0+(9+7-(1)))))-(7)-(4)))-(9))))-7))))+(9))+(10))-(8-(9)))+(8))-(6)-(4)-(8)))))))))))))-(7)))))+(2-(6)-(0))))-(0)-(5+(9)+(9))+(3-(9))))+(8))))))))-(0-(0))+(7-(2))))))))-(6))-(8+(9))-(9+(2))-(2)+(9))-(4))+(7)-(1)-(6))-(2-0)))))))-(0)))))-(8+(0-(5))))+(9)-(1-(0)-(3)))-(3)-(0)))+(4)+(6))))-(5)+(1-(5)))))+(10))))-(5)+(0))))))-(6)))))))+(1))))))))-(5)))))))))+(8))))))))))))))))))-(7)+(10)))))))))))))-(4))))))-(10)-(4))+(1)+(3))-(1))))+(9))))))))+(2-(7-(4-(3+(0))))-(10)))))+0))))+(10)))))+(4)))))))))))+(3)))))))-(5)))))+(3)))))))))))))-(7)-(5-(2+(9))-(0))+(4)))+(10)))))-(1)))-(0))+(1))-(8+(10))))))-(10)-(10+(9)+(2))))-(1)))-(2))))+(4+(5))))))+(8))))))))))))))))))))))-(7)))-(3)))))))+(1))))-(7)-(3)+(4))))))-(6)))))))-(9-(3)))))))))))+(8))))))))))+(6))))))))))))))))))))))))))))))))))+(5))+(7))))))))))))))))))))))))-(10))))))+9)))))))"