Infix to Prefix conversion using two stacks
Infix : An expression is called the Infix expression if the operator appears in between the operands in the expression. Simply of the form (operand1 operator operand2).
Example : (A+B) * (C-D)
Prefix : An expression is called the prefix expression if the operator appears in the expression before the operands. Simply of the form (operator operand1 operand2).
Example : *+AB-CD (Infix : (A+B) * (C-D) )
Given an Infix expression, convert it into a Prefix expression using two stacks.
Examples:
Input : A * B + C / D
Output : + * A B/ C D Input : (A - B/C) * (A/K-L)
Output : *-A/BC-/AKL
分析:
- Traverse the infix expression and check if given character is an operator or an operand.
- If it is an operand, then push it into operand stack.
- If it is an operator, then check if priority of current operator is greater than or less than or equal to the operator at top of the stack. If priority is greater, then push operator into operator stack. Otherwise pop two operands from operand stack, pop operator from operator stack and push string operator + operand1 + operand 2 into operand stack. Keep popping from both stacks and pushing result into operand stack until priority of current operator is less than or equal to operator at top of the operator stack.
- If current character is ‘(‘, then push it into operator stack.
- If current character is ‘)’, then check if top of operator stack is opening bracket or not. If not pop two operands from operand stack, pop operator from operator stack and push string operator + operand1 + operand 2 into operand stack. Keep popping from both stacks and pushing result into operand stack until top of operator stack is an opening bracket.
- The final prefix expression is present at top of operand stack.
class Solution {
boolean isOperator(char C) {
return C == '-' || C == '+' || C == '*' || C == '/' || C == '^';
}
int getPriority(char C) {
if (C == '-' || C == '+') {
return ;
} else if (C == '*' || C == '/') {
return ;
} else if (C == '^') {
return ;
} else {
return ;
}
}
String infixToPrefix(String infix) {
Stack<Character> operators = new Stack<>();
Stack<String> operands = new Stack<String>();
for (int i = ; i < infix.length(); i++) {
if (infix.charAt(i) == '(') {
operators.push(infix.charAt(i));
}
else if (infix.charAt(i) == ')') {
while (!operators.empty() && operators.peek() != '(') {
String op1 = operands.pop();
String op2 = operands.pop();
char op = operators.pop();
String tmp = op + op2 + op1;
operands.push(tmp);
}
} else if (!isOperator(infix.charAt(i))) {
operands.push(infix.charAt(i) + "");
}
else {
while (!operators.empty() && getPriority(infix.charAt(i)) <= getPriority(operators.peek())) {
String op1 = operands.pop();
String op2 = operands.pop();
char op = operators.pop();
operands.push(op + op2 + op1);
}
operators.push(infix.charAt(i));
}
}
while (!operators.empty()) {
String op1 = operands.pop();
String op2 = operands.pop();
char op = operators.pop();
operands.push(op + op2 + op1);
}
return operands.peek();
}
}
Infix to Prefix conversion using two stacks的更多相关文章
- Postfix to Prefix Conversion & Prefix to Postfix Conversion
Postfix to Prefix Conversion Postfix: An expression is called the postfix expression if the operator ...
- Infix to postfix conversion 中缀表达式转换为后缀表达式
Conversion Algorithm 1.操作符栈压入"#": 2.依次读入表达式的每个单词: 3.如果是操作数则压入操作数栈: 4.如果是操作符,则将操作符栈顶元素与要读入的 ...
- Swift声明参考
一条声明可以在你的程序里引入新的名字和构造.举例来说,你可以使用声明来引入函数和方法,变量和常量,或者来定义 新的命名好的枚举,结构,类和协议类型.你也可以使用一条声明来延长一个已经存在的命名好的类型 ...
- Swift5 语言参考(六) 声明
一个声明引入了一个新的名称或构建到你的程序.例如,您使用声明来引入函数和方法,引入变量和常量,以及定义枚举,结构,类和协议类型.您还可以使用声明来扩展现有命名类型的行为,并将符号导入到其他地方声明的程 ...
- a note of R software write Function
Functionals “To become significantly more reliable, code must become more transparent. In particular ...
- SpringBoot @ConfigurationProperties详解
文章目录 简介 添加依赖关系 一个简单的例子 属性嵌套 @ConfigurationProperties和@Bean 属性验证 属性转换 自定义Converter SpringBoot @Config ...
- React 17 要来了,非常特别的一版
写在前面 React 最近发布了v17.0.0-rc.0,距上一个大版本v16.0(发布于 2017/9/27)已经过去近 3 年了 与新特性云集的 React 16及先前的大版本相比,React 1 ...
- Prefix to Infix Conversion
Infix : An expression is called the Infix expression if the operator appears in between the operands ...
- C++ Knowledge series Conversion & Constructor & Destructor
Everything has its lifecycle, from being created to disappearing. Pass by reference instead of pass ...
随机推荐
- has(expr|ele)保留包含特定后代的元素,去掉那些不含有指定后代的元素。
has(expr|ele) 概述 保留包含特定后代的元素,去掉那些不含有指定后代的元素.大理石平台等级 .has()方法将会从给定的jQuery对象中重新创建一组匹配的对象.提供的选择器会一一测试原先 ...
- hdu 1133 卡特兰 高精度
Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...
- [JSOI 2016] 最佳团体(树形背包+01分数规划)
4753: [Jsoi2016]最佳团体 Time Limit: 20 Sec Memory Limit: 512 MBSubmit: 2003 Solved: 790[Submit][Statu ...
- 物联网是前端工程师的新蓝海吗? | Live笔记
物联网是继 Web .无线之后的又一次重大技术变革,在变革的大潮中,程序员的知识体系和思维方式将面临全面更新. 前端开发的历史 在准备这个live的过程中,我回顾了前端开发短暂的历史,有几次我认为非常 ...
- 写简单的tb(testbench)文件来测试之前的FSM控制的LED
先上我之前写的状态机控制的led代码led_test.v module led_test(clk,led_out); input clk; :] led_out; initial begin led_ ...
- CF427D
CF427D SA的奇技淫巧,其实就是板子. 题意: 给定两个字符串,求最短的满足各只出现一次的连续公共字串 解析: 一般情况下,SA都是用来求最长公共前缀的,好像和这道题所求的最短公共子串没有任何关 ...
- linux环境中关闭tomcat,通过shutdown.sh无法彻底关闭--线程池
最近测试环境上测试的项目通过shutdown.sh始终无法彻底关闭. 之前临时解决方法两种: 第一:通过ps -ef|grep tomcat查看到tomcat的进程直接使用kill来杀死进程. 第二: ...
- Qt 单元测试
使用Qtcreator 自带的单元测试工具框架QTestlib进行测试. 一.创建一个单元测试程序 new project->other project ->Qt unit test ...
- [MyBatis]向MySql数据库插入一千万条数据 批量插入用时6分 之前时隐时现的异常不见了
本例代码下载:https://files.cnblogs.com/files/xiandedanteng/InsertMillionComparison20191012.rar 这次实验的环境仍然和上 ...
- 如何用MATLAB GUI创建图形用户界面
MATLAB是众多理工科学生及工程师经常使用的一款数学软件,除了可以实现数据处理,矩阵运算.函数绘制等功能外,MATLAB还可以实现图形用户界面的设计. 下面介绍如何让小白也能用GUI创建最基本的用户 ...