[LeetCode]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.

用堆栈来做，注意处理好字符串的空格，和各种CornerCase。思路不复杂，但是细节要注意。

class Solution {
public:
    int calculate(string s) {
        stack<int> num;
        stack<char> op;
        int front = 0;
        int next = 0; //记录前一个符号和后一个符号
        s = '(' + s + ')';
        op.push('(');
        for(int i=1; i<s.size(); i++){
            if(s[i]==' ') continue; //空格处理跳过
            if(s[i]=='+'||s[i]=='-'||s[i]=='('||s[i]==')'){
                next = i;
                if(!allwhite(s,front,next))//判断两个符号间是否全是空格
                    num.push(atoi(s.substr(front+1,next-front-1).c_str()));//提取数字
                front = next;
                op.push(s[i]);
            }
            if(op.top()==')') // 出栈逻辑
            {
                op.pop();
                int temp = 0;
                while(true)
                {    if(op.top()=='+'){
                    int a = num.top();
                    temp = temp + a;
                    num.pop();
                    op.pop();
                }
                    if(op.top()=='-'){
                        int a = num.top();
                        num.pop();
                        temp = temp - a;
                        op.pop();
                    }
                    if(op.top()=='('){
                        int a = num.top();
                        temp = temp + a;
                        num.pop();
                        num.push(temp);
                        op.pop();
                        break;
                    }
                }
            }
        }
        return num.top();
    }
    bool allwhite(string &s,int i,int j){
        if(j==i)
            return true;
        for(int k=i+1; k<j; k++){
            if(s[k]!=' ')
                return false;
        }
        return true;
    }
};