LeetCode #227 - Basic Calculator II

题目描述：

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.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

• You may assume that the given expression is always valid.
• Do not use the eval built-in library function.

计算一个只包含非负数、空格和加减乘除符号的数学表达式的结果，解题的关键在于定义一个字符记录上一次出现的运算符，当遇到数字时进行判断，如果是加号，直接将数字压栈，如果是负号，将数字的相反数压栈，如果是乘号，将栈顶出栈和当前数值计算乘积再压栈，如果是除号，将栈顶出栈和当前数值计算商再压栈。最终结果为栈中所有数值相加。

class Solution {
public:
    int calculate(string s) {
        stack<int> x;
        int num=0;
        char op='+';
        for(int i=0;i<s.size();i++)
        {
            if(s[i]>='0'&&s[i]<='9') num=num*10+s[i]-'0';
            if(s[i]=='-'||s[i]=='+'||s[i]=='*'||s[i]=='/'||i==s.size()-1)
            {
                if(op=='+') x.push(num);
                else if(op=='-') x.push(-num);
                else
                {
                    int pre=x.top();
                    x.pop();
                    if(op=='*') x.push(pre*num);
                    if(op=='/') x.push(pre/num);
                }
                op=s[i];
                num=0;
            }
        }
        int sum=0;
        while(!x.empty())
        {
            sum+=x.top();
            x.pop();
        }
        return sum;
    }
};