代码语言：javascript复制
今天分享leetcode第20篇文章，也是leetcode第150题—逆波兰表达式求值（Evaluate Reverse Polish Notation），地址是：https://leetcode.com/problems/evaluate-reverse-polish-notation/【英文题目】（学习英语的同时，更能理解题意哟~）Evaluate the value of an arithmetic expression in Reverse Polish Notation.Valid operators are , -, *, /. Each operand may be an integer or another expression.Note:Division between two integers should truncate toward zero.The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.Example 1:Input: ["2", "1", " ", "3", "*"]
Output: 9
Explanation: ((2 1) * 3) = 9
Example 2:Input: ["4", "13", "5", "/", " "]
Output: 6
Explanation: (4 (13 / 5)) = 6
【中文题目】根据逆波兰表示法，求表达式的值。有效的运算符包括 , -, *, / 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。说明：整数除法只保留整数部分。给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。示例 1：输入: ["2", "1", " ", "3", "*"]
输出: 9
解释: ((2 1) * 3) = 9
示例 2：输入: ["4", "13", "5", "/", " "]
输出: 6
解释: (4 (13 / 5)) = 6
【思路】逆波兰表示法，也称为后缀表示法，即将运算符写在操作数之后。比如：3 5的逆波兰表示法为["3", "5", " "]，(3 5) * 2的逆波兰表示法为["3", "5", " ", "2", "*"]。逆波兰式在计算机看来是比较简单易懂的结构，因为计算机普遍采用的内存结构是栈式结构。这是使用栈的典型题，遇到数字添加元素，遇到操作符弹出元素进行相应操作即可。有一点需要注意的是：c 对于两个整数相除结果，正数向下取整，负数向上取整；而python都是向下取整。本题取整需要按照c 取整方式。【代码】python版本class Solution(object):
def evalRPN(self, tokens):
"""
:type tokens: List[str]
:rtype: int
"""
op = set(' -*/')
ls = []
for t in tokens:
# 是符号
if t in op:
b = ls.pop()
a = ls.pop()
if t == ' ':
ls.append(a b)
elif t == '-':
ls.append(a - b)
elif t == '*':
ls.append(a * b)
else:
# python和C 对整数相除取整的不同之处
# c ：正数向下取整，负数向上取整
res = abs(a) / abs(b)
if a * b < 0:
res *= -1
ls.append(res)
print(ls[-1])
# 是数字
else:
ls.append(int(t))
return ls[0]
C 版本class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> num;
set<string> op = {" ", "-", "*", "/"};
int a, b;
for(int i=0; i < tokens.size(); i ){
string t = tokens[i];
if(op.find(t) != op.end()){
b = num.top();
num.pop();
a = num.top();
num.pop();
if(t == " ")
num.push(a b);
if(t == "-")
num.push(a - b);
if(t == "*")
num.push(a * b);
if(t == "/")
num.push(a / b);
}else
num.push(atoi(t.c_str()));
}
return num.top();
}
};
