【LeetCode 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

思路：

　　将中缀表达式转化为后缀表达式然后求值。但是这道题没有括号，所以实现起来相对简单，也有比较简便的方法，就是第一遍先处理乘除法，然后第二遍处理加减法，这样效率高一点。抱着学习如何将中缀表达式转化为后缀表达式的态度，还是用后缀表达式来做的 - -。

C++：

  1 class Solution {
  2 public:
  3
  4     //构造函数
  5     Solution()
  6     {
  7         //设置运算的优先级
  8         priorityMap.insert(make_pair(‘+‘, 0));
  9         priorityMap.insert(make_pair(‘-‘, 0));
 10         priorityMap.insert(make_pair(‘*‘, 1));
 11         priorityMap.insert(make_pair(‘/‘, 1));
 12     }
 13
 14     //中缀表达式转换为后缀表达式
 15     void inFix2PostFix(string& s, vector<int>& PostFix)
 16     {
 17         stack<char> stk;
 18
 19         for(int i = 0; i < s.size(); i++)
 20         {
 21             if(s[i] >= ‘0‘ && s[i] <= ‘9‘)
 22             {
 23                 int sum = s[i++] - ‘0‘;
 24
 25                 while(s[i] >= ‘0‘ && s[i] <= ‘9‘)
 26                 {
 27                     sum = sum*10 + s[i] - ‘0‘;
 28                     i++;
 29                 }
 30
 31                 PostFix.push_back(sum);
 32
 33                 i--;
 34             }
 35
 36             if(s[i] == ‘+‘ || s[i] == ‘-‘ || s[i] == ‘*‘ || s[i] == ‘/‘)
 37             {
 38                 while(!stk.empty())
 39                 {
 40                     char topChar = stk.top();
 41                     if(priorityMap[topChar] >= priorityMap[s[i]])
 42                     {
 43                         PostFix.push_back(topChar - 48);
 44                         stk.pop();
 45                     }
 46                     else
 47                     {
 48                         break;
 49                     }
 50                 }
 51                 stk.push(s[i]);
 52             }
 53         }
 54
 55         while(!stk.empty())
 56         {
 57             PostFix.push_back(stk.top() - 48);
 58             stk.pop();
 59         }
 60     }
 61
 62     //计算后缀表达式的值
 63     int calculatePostFix(vector<int>& PostFix)
 64     {
 65         stack<int> stk;
 66
 67         for(int i = 0; i < PostFix.size(); i++)
 68         {
 69             if(PostFix[i] >= 0)
 70             {
 71                 stk.push(PostFix[i]);
 72                 continue;
 73             }
 74
 75             int rhs = stk.top();//右操作数
 76             stk.pop();
 77             int lhs = stk.top();//左操作数
 78             stk.pop();
 79
 80             if(PostFix[i] == -5) //加
 81             {
 82                 stk.push(lhs + rhs);
 83             }
 84             if(PostFix[i] == -3)//减
 85             {
 86                 stk.push(lhs - rhs);
 87             }
 88             if(PostFix[i] == -6)//乘
 89             {
 90                 stk.push(lhs * rhs);
 91             }
 92             if(PostFix[i] == -1)//除
 93             {
 94                 stk.push(lhs / rhs);
 95             }
 96         }
 97
 98         return  stk.top();
 99     }
100
101     int calculate(string s) {
102         int len = s.size();
103         if(len == 0)
104             return 0;
105
106         vector<int> PostFix;
107
108         inFix2PostFix(s, PostFix);
109
110         return calculatePostFix(PostFix);
111     }
112 private:
113
114     map<char, int> priorityMap;
115 };