标签：
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.
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.
两个要点：
1、无括号时，顺序执行
2、有括号时，先执行括号中的
两个栈：
一个存放操作数，每次进栈要注意，如果操作符栈顶元素为‘+‘/‘-‘，则需要立即计算。
一个存放操作符（包括括号），每次出现‘)‘时，不断进行出栈计算再进栈，直到弹出‘(‘，说明当前括号内计算完毕。
class Solution { public: int calculate(string s) { stack<int> num; stack<int> op; int i = 0; while(i < s.size()) { while(i < s.size() && s[i] == ‘ ‘) i ++; if(i == s.size()) break; if(s[i] == ‘+‘ || s[i] == ‘-‘ || s[i] == ‘(‘) { op.push(s[i]); i ++; } else if(s[i] == ‘)‘) { while(op.top() != ‘(‘) {// calculation within parentheses int n2 = num.top(); num.pop(); int n1 = num.top(); num.pop(); if(op.top() == ‘+‘) num.push(n1 + n2); else num.push(n1 - n2); op.pop(); } op.pop(); while(!op.empty() && op.top() != ‘(‘) { int n2 = num.top(); num.pop(); int n1 = num.top(); num.pop(); if(op.top() == ‘+‘) num.push(n1 + n2); else num.push(n1 - n2); op.pop(); } i ++; } else { int n = 0; while(i < s.size() && s[i] >= ‘0‘ && s[i] <= ‘9‘) { n = n*10 + (s[i]-‘0‘); i ++; } num.push(n); while(!op.empty() && op.top() != ‘(‘) { int n2 = num.top(); num.pop(); int n1 = num.top(); num.pop(); if(op.top() == ‘+‘) num.push(n1 + n2); else num.push(n1 - n2); op.pop(); } } } return num.top(); } };
【LeetCode】224. Basic Calculator
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原文地址：http://www.cnblogs.com/ganganloveu/p/4640589.html
