227. Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:

“3+2*2” = 7
” 3/2 ” = 1
” 3+5 / 2 ” = 5

Note: Do not use the eval built-in library function.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

设计一个简单的计算器，只涉及非负整数的运算，只有加减乘除运算。我使用栈来实现。遇到数字就解析出当前的数，如果最近的运算号是乘号*或除号/就将前一个数和当前的数结合然后入栈，如果是减号-就将当前的数乘-1再入栈，其他情况直接入栈。最后将栈中所有的数累加即可。

代码：

class Solution{public:    int calculate(string s)     {        stack<int> nums;        char oper = '+';        int i = 0;        s += '$';        while(s[i] != '$')        {            while(s[i] == ' ') ++i;            if(s[i] >= '0' && s[i] <= '9')            {                int num = 0;                while(s[i] >= '0' && s[i] <= '9')                {                    num = num * 10 + (s[i] - '0');                    ++i;                }if(oper == '*')                {                    int tmp = nums.top();                    nums.pop();                    nums.push(tmp * num);                }                else if(oper == '/')                {                    int tmp = nums.top();                    nums.pop();                    nums.push(tmp / num);                }                else if(oper == '-')                {                    nums.push(-1 * num);                }                else                {                    nums.push(num);                }            }            else if(s[i] != '$')            {                oper = s[i++];            }        }        int res = 0;        while(!nums.empty())        {            res += nums.top();            nums.pop();        }        return res;    }};