leetcode_Basic Calculator

题目：

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.

思路：栈的使用，括号的优先级最高。

class Solution {
public:
    int calculate(string s) {
        int len = s.length();
	stack<int> st;
	int i = 0;
	int result = 0;
	while(i<len)
	{
		int sum = 0;
		if(s.at(i)>='0'&&s.at(i)<='9')
		{
			int j = i+1;
			sum = s.at(i)-'0';
			while(j<=len-1&&s.at(j)>='0'&&s.at(j)<='9')
			{
				sum = (sum*10 + (s.at(j) - '0'));
				j++;
			}
		//	cout<<sum<<endl;
			//以上计算数字字符串转化为数字
			if(!st.empty()&&(char)st.top()=='+')
			{
				st.pop();
				result = st.top()+sum;
				st.pop();
				st.push(result);
			}
			else if(!st.empty()&&(char)st.top()=='-')
			{
				st.pop();
				result = st.top()-sum;
				st.pop();
				st.push(result);
			}
			else
			{
				st.push(sum);
			}
			i = j;
		}
 
		else if(s.at(i)==' ')
		{
			i++;
		}
		else if(s.at(i)=='+'||s.at(i)=='-')
		{
			st.push((int)s.at(i));
			i++;
		}
		else if(s.at(i)=='(')
		{
			st.push((int)s.at(i));
			i++;
		}
		else if(s.at(i)==')')
		{
			int temp = st.top();
			if(!st.empty())
			st.pop();
			if(!st.empty())
			st.pop();
			if(!st.empty()&&st.top()=='+')
			{
				st.pop();//去掉
				temp = temp+(st.top());
				st.pop();
				st.push(temp);
			}
			else if(!st.empty()&&st.top()=='-')
			{
				st.pop();//去掉
				temp = (st.top())-temp;
				st.pop();
				st.push(temp);
			}
			else
			{
				st.push(temp);
			}
		    i++;
		}
	}
	return st.top();
    }
};