【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;

     }

 };

巴特西

【LeetCode】227. Basic Calculator

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