我正在尝试编写一个找到中缀的计算程序。此外，用户将输入x变量的数字，程序将解决它。我的程序有效，但它只是第一次解决它。以下时间它给出了与第一次相同的答案。
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;
class Stack {
char a[] = new char[100];
int top = -1;
void push(char c) {
try {
a[++top] = c;
} catch (StringIndexOutOfBoundsException e) {
System.out.println("Stack full , no room to push , size=100");
System.exit(0);
}
}
char pop() {
return a[top--];
}
boolean isEmpty() {
return (top == -1) ? true : false;
}
char peek() {
return a[top];
}
}
public class intopost {
static Stack operators = new Stack();
public static void main(String argv[]) throws IOException {
String infix;
// create an input stream object
BufferedReader keyboard = new BufferedReader(new InputStreamReader(
System.in));
// get input from user
System.out.print("\nEnter the algebraic expression in infix: ");
infix = keyboard.readLine();
String postFx = toPostfix(infix);
// output as postfix
System.out.println("The expression in postfix is:" + postFx);
if (postFx.contains("x")) {
String line = "";
do {
System.out.println("Enter value of X : ");
line = keyboard.readLine();
if (!"q".equalsIgnoreCase(line)) {
postFx = postFx.replaceAll("x", line);
System.out.println("Answer to expression : "
+ EvaluateString.evaluate(postFx));
}
} while (!line.equals("q"));
} else {
System.out.println("Answer to expression : "
+ EvaluateString.evaluate(postFx));
}
}
private static String toPostfix(String infix)
// converts an infix expression to postfix
{
char symbol;
String postfix = "";
for (int i = 0; i < infix.length(); ++i)
// while there is input to be read
{
symbol = infix.charAt(i);
// if it's an operand, add it to the string
if (symbol != ' ') {
if (Character.isLetter(symbol) || Character.isDigit(symbol))
postfix = postfix + " " + symbol;
else if (symbol == '(')
// push (
{
operators.push(symbol);
} else if (symbol == ')')
// push everything back to (
{
while (operators.peek() != '(') {
postfix = postfix + " " + operators.pop();
}
operators.pop(); // remove '('
} else
// print operators occurring before it that have greater
// precedence
{
while (!operators.isEmpty() && !(operators.peek() == '(')
&& prec(symbol) <= prec(operators.peek()))
postfix = postfix + " " + operators.pop();
operators.push(symbol);
}
}
}
while (!operators.isEmpty())
postfix = postfix + " " + operators.pop();
return postfix.trim();
}
static int prec(char x) {
if (x == '+' || x == '-')
return 1;
if (x == '*' || x == '/' || x == '%')
return 2;
return 0;
}
}
class EvaluateString {
public static int evaluate(String expression) {
char[] tokens = expression.toCharArray();
// Stack for numbers: 'values'
LinkedList<Integer> values = new LinkedList<Integer>();
// Stack for Operators: 'ops'
LinkedList<Character> ops = new LinkedList<Character>();
for (int i = 0; i < tokens.length; i++) {
// Current token is a whitespace, skip it
if (tokens[i] == ' ')
continue;
// Current token is a number, push it to stack for numbers
if (tokens[i] >= '0' && tokens[i] <= '9') {
StringBuffer sbuf = new StringBuffer();
// There may be more than one digits in number
while (i < tokens.length && tokens[i] >= '0'
&& tokens[i] <= '9')
sbuf.append(tokens[i++]);
values.push(Integer.parseInt(sbuf.toString()));
}
// Current token is an opening brace, push it to 'ops'
else if (tokens[i] == '(')
ops.push(tokens[i]);
// Closing brace encountered, solve entire brace
else if (tokens[i] == ')') {
while (ops.peek() != '(')
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
ops.pop();
}
// Current token is an operator.
else if (tokens[i] == '+' || tokens[i] == '-' || tokens[i] == '*'
|| tokens[i] == '/') {
// While top of 'ops' has same or greater precedence to current
// token, which is an operator. Apply operator on top of 'ops'
// to top two elements in values stack
while (!ops.isEmpty() && hasPrecedence(tokens[i], ops.peek()))
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
// Push current token to 'ops'.
ops.push(tokens[i]);
}
}
// Entire expression has been parsed at this point, apply remaining
// ops to remaining values
while (!ops.isEmpty())
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
// Top of 'values' contains result, return it
return values.pop();
}
// Returns true if 'op2' has higher or same precedence as 'op1',
// otherwise returns false.
public static boolean hasPrecedence(char op1, char op2) {
if (op2 == '(' || op2 == ')')
return false;
if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-'))
return false;
else
return true;
}
// A utility method to apply an operator 'op' on operands 'a'
// and 'b'. Return the result.
public static int applyOp(char op, int b, int a) {
switch (op) {
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (b == 0)
throw new UnsupportedOperationException("Cannot divide by zero");
return a / b;
}
return 0;
}
答案 0 :(得分：0)
main方法中的循环内部存在错误。请参阅下面的代码段。
postFx = postFx.replaceAll("x", line);
System.out.println("Answer to expression : "
+ EvaluateString.evaluate(postFx));
此处postFx = postFx.replaceAll("x", line);您丢失了对包含变量x的postfix表单的引用。后续调用replaceAll没有任何效果。因此，评估具有第一个输入值的表达式。
您可以通过用
System.out.println("Answer to expression : "
+ EvaluateString.evaluate(postFx.replaceAll("x", line)));