版权声明:本文为博主NJU_ChopinXBP原创文章,发表于CSDN,仅供交流学习使用,转载请私信或评论联系,未经博主允许不得转载。感谢您的评论与点赞。 https://blog.csdn.net/qq_20304723/article/details/89102288
2019.4.8
开始学习使用markdown编辑器进行编辑,感觉还挺顺手。
后缀表达式在合法的情况下,运算过程实际上只要借助一个栈进行。
当字符为数字时,元素入栈;当字符为符号时,栈顶元素出栈并进行运算。
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.
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。
扫描二维码关注公众号,回复:
5816974 查看本文章
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
示例 3:
输入: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
输出: 22
解释:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22
/**
*
* @author ChopinXBP
* Evaluate the value of an arithmetic expression in Reverse Polish Notation.
* Valid operators are+,-,*,/. Each operand may be an integer or another expression.
* 模拟后缀表达式
*
*
*/
import java.util.Stack;
public class EvaluateReversePolishNotation {
public static String[] str = {"2", "1", "+", "3", "*"};
public static String[] str1 = {"0", "3", "/"};
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println(evalRPN(str1));
}
public static int evalRPN(String[] tokens) {
Stack<Integer> element = new Stack<Integer>();
for(int i = 0; i < tokens.length; i++){
if(tokens[i].equals("+")){
int right = element.pop();
int left = element.pop();
element.push(left + right);
}
else if(tokens[i].equals("-")){
int right = element.pop();
int left = element.pop();
element.push(left - right);
}
else if(tokens[i].equals("*")){
int right = element.pop();
int left = element.pop();
element.push(left * right);
}
else if(tokens[i].equals("/")){
int right = element.pop();
int left = element.pop();
element.push(left / right);
}
else{
element.push(Integer.parseInt(tokens[i]));
}
}
return element.pop();
}
}
#Coding一小时,Copying一秒钟。留个言点个赞呗,谢谢你#