【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" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

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

 1 class Solution {
 2 public:
 3     int calculate(string s)
 4     {
 5        int res=0, num=0, op=‘+‘, i=0, mid=0;
 6        while(i<s.size())
 7        {
 8            num = 0;  ////每一轮要初始化num
 9            //获取一整个数字存储在num中
10            if(isdigit(s[i]))
11            {
12                 while(i<s.size()&&isdigit(s[i]))  //连续数字字符作一个数
13                 {
14                      num = num*10+(s[i]-‘0‘);
15                      i++;
16                 }
17
18                //获取运算符存储在op中
19                 if(op==‘+‘||op==‘-‘)
20                 {
21                     res = res+mid;
22                     mid = num*(op==‘-‘?-1:1);
23                 }
24                 else if(op==‘/‘)
25                 {
26                     mid = mid/num;
27                 }
28                 else if(op==‘*‘)
29                 {
30                     mid = mid*num;
31                 }
32            }
33            else if(s[i]==‘ ‘)         //不要漏了开头空格这种情况= - =它喵被摆了一道
34                 i++;
35            else
36            {
37                 op=s[i];
38                 i++;
39            }
40        }
41         res = res+mid;
42         return res;
43     }
44 };