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
一道混合的题，难度不高，是几个部分组成。
需要注意：
1.转换数字，负数的存在。
2.出栈时数字的顺序。
3.好像没什么了……
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57 |
class
Solution {public: int
evalRPN(vector<string> &tokens) { stack<int> sk; int
result; for(int
i = 0 ; i < tokens.size(); i++) { if(isop(tokens[i])) { int
second = sk.top(); sk.pop(); int
first = sk.top(); sk.pop(); int
midresult; if(tokens[i] == "+") midresult = first + second; else
if(tokens[i] == "-")midresult = first - second; else
if(tokens[i] == "*")midresult = first * second; else
if(tokens[i] == "/")midresult = first / second; sk.push(midresult); } else { int
tp = string2num(tokens[i]); sk.push(tp); } } return
sk.top(); } bool
isop(string s) { if(s== "+"
||s== "-" ||s== "*" ||s== "/"
)return
1; else
return 0; } int
string2num(string s) { int
re =0 ; int
flag = 0; int
i = 0; if(s[0] == ‘-‘) { flag =1; i++; } for( ; i < s.size();i++) { re = re*10; re = re + s[i]-‘0‘; } if(flag ==1) re = 0 -re; return
re; }}; |
Evaluate Reverse Polish Notation,布布扣,bubuko.com
Evaluate Reverse Polish Notation
原文：http://www.cnblogs.com/pengyu2003/p/3621164.html