Evaluate Reverse Polish Notation 计算逆波兰表达式

　Evaluate the value of an arithmetic expression in Reverse Polish Notation.

　　Valid operators are +, -, *, /. Each operand may be an integer or another expression.

　　Some examples:   ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9   ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

  • 题目大意：给定一个逆波兰表达式，求该表达式的值
  • 思路：由于逆波兰表达式本身不需要括号来限制哪个运算该先进行，因此可以直接利用栈来模拟计算：遇到操作数直接压栈，碰到操作符直接取栈顶的2个操作数进行计算（注意第一次取出来的是右操作数），然后再把计算结果压栈，如此循环下去。最后栈中剩下的唯一一个元素便是整个表达式的值
  • 实现： 
     1 class Solution {
 2 public:
 3     int evalRPN(vector<string> &tokens) {
 4          
 5         int result = 0;
 6         int i;
 7         stack<int> opd;         //存储操作数
 8         int size = tokens.size();
 9         for(i=0;i<size;i++)
10         {
11             if(tokens[i]=="*")
12             {
13                 int rOpd = opd.top();   //右操作数
14                 opd.pop();
15                 int lOpd = opd.top();  //左操作数
16                 opd.pop();
17                 result = lOpd*rOpd;
18                 opd.push(result);
19             }
20             else if(tokens[i]=="/")
21             {
22                 int rOpd = opd.top();
23                 opd.pop();
24                 int lOpd = opd.top();
25                 opd.pop();
26                 result = lOpd/rOpd;
27                 opd.push(result);
28             }
29             else if(tokens[i]=="+")
30             {
31                 int rOpd = opd.top();
32                 opd.pop();
33                 int lOpd = opd.top();
34                 opd.pop();
35                 result = lOpd+rOpd;
36                 opd.push(result);
37             }
38             else if(tokens[i]=="-")
39             {
40                 int rOpd = opd.top();
41                 opd.pop();
42                 int lOpd = opd.top();
43                 opd.pop();
44                 result = lOpd-rOpd;
45                 opd.push(result);
46             }
47             else
48             {
49                 opd.push(atoi(tokens[i].c_str()));
50             }
51         }
52         return opd.top();
53     }
54 };

     

class Solution {
public:
    int evalRPN(vector<string> &tokens) {
         
        int result = 0;
        int i;
        stack<int> opd;         //存储操作数
        int size = tokens.size();
        for(i=0;i<size;i++)
        {
            if(tokens[i]=="*")
            {
                int rOpd = opd.top();   //右操作数
                opd.pop();
                int lOpd = opd.top();  //左操作数
                opd.pop();
                result = lOpd*rOpd;
                opd.push(result);
            }
            else if(tokens[i]=="/")
            {
                int rOpd = opd.top();
                opd.pop();
                int lOpd = opd.top();
                opd.pop();
                result = lOpd/rOpd;
                opd.push(result);
            }
            else if(tokens[i]=="+")
            {
                int rOpd = opd.top();
                opd.pop();
                int lOpd = opd.top();
                opd.pop();
                result = lOpd+rOpd;
                opd.push(result);
            }
            else if(tokens[i]=="-")
            {
                int rOpd = opd.top();
                opd.pop();
                int lOpd = opd.top();
                opd.pop();
                result = lOpd-rOpd;
                opd.push(result);
            }
            else
            {
                opd.push(atoi(tokens[i].c_str()));
            }
        }
        return opd.top();
    }
};
posted @ 2020-09-04 17:04  cuanduan肝颤  阅读(144)  评论(0)    收藏  举报