leetcode栈--1、evaluate-reverse-polish-notation（逆波兰表达式）

题目描述

Evaluate the value of an arithmetic expression inReverse 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

解题思路：使用栈结构来存储。遇到数字存入栈中，遇到符号，弹出栈顶的两个元素，进行操作，并将结果放入栈中。

 1 class Solution {
 2 public:
 3     int strtoint(string s) // string转int
 4     {
 5         int result = 0;
 6         int base = 1;//记录属于的位
 7         int t = 1;//记录正负号
 8         if(s[0] == ‘-‘)
 9             t = -1;
10         for(int i= s.size() - 1;i>=0;i--)
11         {
12             if(s[i]>=‘0‘ && s[i]<=‘9‘)//此处一定注意0和9加‘‘
13             {
14                 result += base * (s[i] - ‘0‘);
15                 base *= 10;
16             }
17         }
18         return result*t;
19     }
20     int evalRPN(vector<string> &tokens) {
21         int n = tokens.size();
22         if(tokens.empty() || n<=0)
23             return 0;
24         stack<int> s;
25         for(int i=0;i<n;i++)
26         {
27             if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/")
28             {
29                 int num2 = s.top();//先弹出的为右操作数
30                 s.pop();
31                 int num1 = s.top();
32                 s.pop();
33                 if(tokens[i] == "+")
34                 {
35                     s.push(num1+num2);
36                 }
37                 else if(tokens[i] == "-")
38                 {
39                     s.push(num1-num2);
40                 }
41                 else if(tokens[i] == "*")
42                 {
43                     s.push(num1*num2);
44                 }
45                 else
46                 {
47                     s.push(num1/num2);
48                 }
49             }
50             else
51             {
52                 s.push(strtoint(tokens[i]));
53             }
54         }
55         return s.top();
56     }
57 };