【LeetCode】227. Basic Calculator

Problem:

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" =
" 3/2 " =
" 3+5 / 2 " = 

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

Solution:

字符串分成两种情况：数字num和运算符op，结果设为res，考虑到先乘除后加减的运算法则，设中间数mid优先计算乘除，再与res相加.

*> C++ version

 class Solution {
 public:
     int calculate(string s)
     {
        int res=, num=, op='+', i=, mid=;
        while(i<s.size())
        {
            num = ;  ////每一轮要初始化num
            //获取一整个数字存储在num中
            if(isdigit(s[i]))
            {
                 while(i<s.size()&&isdigit(s[i]))  //连续数字字符作一个数
                 {
                      num = num*+(s[i]-'');
                      i++;
                 }

                //获取运算符存储在op中
                 if(op=='+'||op=='-')
                 {
                     res = res+mid;
                     mid = num*(op=='-'?-:);
                 }
                 else if(op=='/')
                 {
                     mid = mid/num;
                 }
                 else if(op=='*')
                 {
                     mid = mid*num;
                 }
            }
            else if(s[i]==' ')         //不要漏了开头空格这种情况= - =它喵被摆了一道
                 i++;
            else
            {
                 op=s[i];
                 i++;
            }
        }
         res = res+mid;
         return res;
     }
 };