Java 将后缀表达式转换为中缀,计算后缀并给出答案
我一直在做这个程序,接收一个后缀表达式,计算并打印它的答案,还把它转换成中缀表达式。我已经让它的计算部分工作了。例如,如果我输入2+(考虑间距),它会给我4。但是,对于中缀部分,它打印出2+。然后我试了一下,没有任何间隔。我投了22+。后缀不起作用,但中缀打印正确。所以我不知道如何修复这部分。这是我的密码Java 将后缀表达式转换为中缀,计算后缀并给出答案,java,tree,postfix-notation,Java,Tree,Postfix Notation,我一直在做这个程序,接收一个后缀表达式,计算并打印它的答案,还把它转换成中缀表达式。我已经让它的计算部分工作了。例如,如果我输入2+(考虑间距),它会给我4。但是,对于中缀部分,它打印出2+。然后我试了一下,没有任何间隔。我投了22+。后缀不起作用,但中缀打印正确。所以我不知道如何修复这部分。这是我的密码 import java.util.NoSuchElementException; import java.util.Stack; public class ExpressionTree
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 later its used to construct the expression tree.
* The posfix expression, if invalid, leads to invalid results
*
* @param postfix the postfix expression.
*/
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;
}
}
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
{
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();
}
}
任何帮助都将不胜感激,因为我真的不知道如何解决这个问题,我是如此接近完成它
if (isOperator(ch))
{
TreeNode rightNode = nodes.pop();
TreeNode leftNode = nodes.pop();
nodes.push(new TreeNode(leftNode, ch, rightNode));
}
else
{
nodes.add(new TreeNode(null, ch, null));
}
您正在将空白节点放入树中。试试这个:
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));
}
非常感谢。这解决了问题!
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));
}