我不确定我是否正确行事。我正在尝试将中缀方程转换为后缀方程，如：
(3 + 4) * 2
是：
4 3 + 2 *
如果可能，我试图在一种方法中完成所有这些。
现在我收到了一个arrayoutofbounds错误，所以我在错误的地方或其他地方弹出。
这是infixtopostfix方法。
public void InfixToPostfix(String f) {
Stacked postfix = new Stacked(100);
Stacked op = new Stacked(100);
char c;
String fix = f;
//String out = "";
int out = 0;
for (int i = 0; i < fix.length(); i++) {
c = fix.charAt(i);
if (c != '+' || c != '*' || c != '-' || c != '/' || c != '(' || c != ')') {
// out += c;
postfix.push(c);
}
else if (c == '+' || c == '*' || c == '-' || c == '/') {
if (c != ')') {
op.push(c);
} else {
out = (char) op.pop();
s.push(out);
}
out = (char) op.pop();
System.out.println("THE POSTFIX = " + out)
}
}
}
答案 0 :(得分：3)
public class Postfix {
private int priority(char ch) {
if (ch == '^')
return 3;
if (ch == '/' || ch == '*')
return 2;
if (ch == '+' || ch == '-')
return 1;
return 0;
}
public String toPostfix(String in ) {
String copy = in +")";
Stack s = new Stack(copy.length());
s.push('(');
int i, l = copy.length();
char ch;
String r = "";
for (i = 0; i < l; i++) {
ch = copy.charAt(i);
if (Character.isLetter(ch) == true)
r += ch;
else if (ch == '(')
s.push(ch);
else if (ch == ')') {
while (s.seeTop() != '(')
r += s.popSeeTop();
s.pop();
} else {
while (priority(ch) <= priority(s.seeTop()))
r += s.popSeeTop();
s.push(ch);
}
}
return r;
}
}
答案 1 :(得分：2)
public class InfixToPostfix {
Stack operatorStack = new Stack();
public String toPrefix(String expression) {
expression="("+expression+")";
int i;
char token;
String output = "";
for (i = 0; i < expression.length(); i++) {
token = expression.charAt(i);
if (Character.isLetterOrDigit(token) == true)
output += token;
else if (token == '(')
operatorStack.push(token);
else if (token == ')') {
char seeTop;
while ((seeTop = seeTop()) != '(') {
output += seeTop;
popSeeTop();
}
operatorStack.pop();
} else {
while (priority(token) <= priority(seeTop())) {
output += seeTop();
popSeeTop();
}
operatorStack.push(token);
}
}
return output;
}
private int priority(char operator) {
if (operator == '^')
return 3;
if (operator == '/' || operator == '*')
return 2;
if (operator == '+' || operator == '-')
return 1;
return 0;
}
private Character seeTop() {
if(!operatorStack.empty())
return (Character) operatorStack.peek();
else
return '0';
}
private void popSeeTop() {
if(!operatorStack.empty())
operatorStack.pop();
}
答案 2 :(得分：-1)
/**
* Convert arithmetic from infix to postfix
*
* @example 1 * ( 12 + 23 ) - ( 4 / 5 ) => 1 12 23 + * 4 5 / -
* @author Yong Su
*/
import java.util.Stack;
import java.lang.StringBuilder;
class InfixToPostfix {
public static void main(String[] args) {
String infix = "1 * ( 12 + 23 ) - ( 4 / 5 )";
String postfix = convert(infix);
System.out.println(postfix);
}
/**
* Perform the conversion
*
* @param expression Arithmetic infix format
*/
public static String convert(String expression) {
// Use StringBuilder to append string is faster than
// String concatenation
StringBuilder sb = new StringBuilder();
// Use a stack to track operations
Stack<Character> op = new Stack<Character>();
// Convert expression string to char array
char[] chars = expression.toCharArray();
// The length of expression character
int N = chars.length;
for (int i = 0; i < N; i++) {
char ch = chars[i];
if (Character.isDigit(ch)) {
// Number, simply append to the result
while (Character.isDigit(chars[i])) {
sb.append(chars[i++]);
}
// Use space as delimiter
sb.append(' ');
} else if (ch == '(') {
// Left parenthesis, push to the stack
op.push(ch);
} else if (ch == ')') {
// Right parenthesis, pop and append to the result until meet the left parenthesis
while (op.peek() != '(') {
sb.append(op.pop()).append(' ');
}
// Don't forget to pop the left parenthesis out
op.pop();
} else if (isOperator(ch)) {
// Operator, pop out all higher priority operators first and then push it to the stack
while (!op.isEmpty() && priority(op.peek()) >= priority(ch)) {
sb.append(op.pop()).append(' ');
}
op.push(ch);
}
}
// Finally pop out whatever left in the stack and append to the result
while(!op.isEmpty()) {
sb.append(op.pop()).append(' ');
}
return sb.toString();
}
/**
* Check if a character is an operator
*/
private static boolean isOperator(char ch) {
return ch == '^' || ch == '*' || ch == '/' || ch == '+' || ch == '-';
}
/**
* Define the operator priority
*/
private static int priority(char operator) {
switch (operator) {
case '^' : return 3;
case '*' :
case '/' : return 2;
case '+' :
case '-' : return 1;
}
return 0;
}
}
答案 3 :(得分：-1)
使用简单的Stacks
public class InfixToPostfix {
public static void main(String[] args) {
Stack<String> stack = new Stack<String>();
while (!StdIn.isEmpty()) {
String s = StdIn.readString();
if (s.equals("+")) stack.push(s);
else if (s.equals("*")) stack.push(s);
else if (s.equals(")")) StdOut.print(stack.pop() + " ");
else if (s.equals("(")) StdOut.print("");
else StdOut.print(s + " ");
}
StdOut.println();
}
}