Question

java后缀表达式实现表达式求值

Answer 1

import java.util.Scanner;
import java.util.Stack;

public class 表达式计算 {
private static Stack<String> num = new Stack<String>();//存后缀表达式
private static Stack<String> sign = new Stack<String>();//存入符号
private static Stack<Integer> result = new Stack<Integer>();//放结果

public static void getGroup(String line){//讲字符串转换为后缀表达式
for(int i=0; i<line.length(); i++){
char c = line.charAt(i);
if((int)c>=48 && (int)c<=57){//当遇到数字的时候,判断是不是多位数，然后在push进num
int j = i+1;
while(j<line.length() && (line.charAt(j)>=48 && line.charAt(j)<=57)){
j++;
}
num.push(line.substring(i, j));
i = j-1;
}else if(c == '('){//遇到左括号直接存进num
sign.push(String.valueOf(c));
}else if(c == ')'){//遇到右括号从sign中pop栈顶元素push到num知道遇到'(',然后再pop掉'('
while(!sign.peek().equals("(")){
num.push(sign.pop());
}
sign.pop();
}else{
int n = 0;
if(!sign.empty()){//如果sign中没有元素，直接令n = 0
n = getNum(sign.peek().charAt(0));
}
int m = getNum(c);
if(m >= n){//如果当前元素的运算级别比栈顶元素运算级别要高，就直接push进sign
sign.push(String.valueOf(c));
}else{
while(m < n){//如果当前运算运算级别比sign栈顶元素运算级别要低，就将sign栈顶元素pop并且push进num，知道不符合条件
num.push(sign.pop());//输入例子2*3+6/3的时候，这里一直报错
if(!sign.empty()){
n = getNum(sign.peek().charAt(0));
}else{
n = 0;
}
}
sign.push(String.valueOf(c));
}
}
}
while(!sign.empty()){
num.push(sign.pop());
}
}

private static int getNum(char c){
int n = 0;
switch(c){
case '+':
case '-':
n = 1;
break;
case '*':
case '/':
n = 2;
break;
}
return n;
}

Answer 2

需要进行语法树解析，如逆波兰表达式