我正在完成一个表达式树程序,它接受一个后缀并以中缀形式给出答案和原始方程。我大约6个月前开始工作,并且遇到了一个我无法弄清楚的问题。如果数字小于9,它工作正常。例如,2 5 * 4 - 打印出中缀:(2 * 5-4)6.0工作正常。类似10 2 *的东西打印出20.0但是打印出中缀:(0 * 2)这是不对的。最大的问题是,如果一个数字大于9,它就不能正确地读取该数字,这会将转换混淆为中缀。我不确定如何解决这个问题。以下是我的班级和测试员班级:
import java.util.NoSuchElementException;
import java.util.Stack;
public class ExpressionTree
{
private final String postfix;
private TreeNode root;
/**
* Takes in a valid postfix expression and its used to construct the expression tree.
*/
public ExpressionTree(String postfix)
{
if (postfix == null) { throw new NullPointerException("The posfix should not be null"); }
if (postfix.length() == 0) { throw new IllegalArgumentException("The postfix should not be empty"); }
this.postfix = postfix;
}
private static class TreeNode
{
TreeNode left;
char ch;
TreeNode right;
TreeNode(TreeNode left, char ch, TreeNode right) {
this.left = left;
this.ch = ch;
this.right = right;
}
}
/**
* Checks to see if it is an Operator
*/
private boolean isOperator(char c) {
return c == '+' || c == '-' || c == '*' || c == '/';
}
/**
* Constructs an expression tree, using the postfix expression
*/
public void createExpressionTree()
{
final Stack<TreeNode> nodes = new Stack<TreeNode>();
for (int i = 0; i < postfix.length(); i++)
{
char ch = postfix.charAt(i);
if (isOperator(ch))
{
TreeNode rightNode = nodes.pop();
TreeNode leftNode = nodes.pop();
nodes.push(new TreeNode(leftNode, ch, rightNode));
}
else if (!Character.isWhitespace(ch))
{
nodes.add(new TreeNode(null, ch, null));
}
}
root = nodes.pop();
}
/**
* Returns the infix expression
*
* @return the string of infix.
*/
public String infix()
{
if (root == null)
{
throw new NoSuchElementException("The root is empty, the tree has not yet been constructed.");
}
final StringBuilder infix = new StringBuilder();
inOrder(root, infix);
return infix.toString();
}
private void inOrder(TreeNode node, StringBuilder infix)
{
if (node != null) {
inOrder(node.left, infix);
infix.append(node.ch);
inOrder(node.right, infix);
}
}
public Double evaluate(String postfix)
{
Stack<Double> s = new Stack<Double>();
char[] chars = postfix.toCharArray();
int N = chars.length;
for(int i = 0; i < N; i++)
{
char ch = chars[i];
if(isOperator(ch))
{
switch(ch)
{
case '+': s.push(s.pop() + s.pop()); break;
case '*': s.push(s.pop() * s.pop()); break;
case '-': s.push(-s.pop() + s.pop()); break;
case '/': s.push(1 / s.pop() * s.pop()); break;
}
}
else if(Character.isDigit(ch))
{
s.push(0.0);
while (Character.isDigit(chars[i]))
s.push(10.0 * s.pop() + (chars[i++] - '0'));
}
}
return s.pop();
}
}
测试人员:
import java.util.Scanner;
public class ExpressionTester
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String line = null;
while(true)
{
System.out.println("");
String pf = sc.nextLine();
if (pf.isEmpty())
{
break;
}
ExpressionTree eT = new ExpressionTree(pf);
eT.createExpressionTree();
System.out.println("The infix: " + "(" + eT.infix() + ")" );
System.out.println(eT.evaluate(pf));
}
}
}