英文:
Alternative way to replace Nested While Loop in Java
问题
我尝试在Java中使用以下代码计算数学表达式：
public double applyOp(char op, double b, double a) {
switch (op) {
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
return a / b;
}
return 0;
}
public boolean hasPrecedence(char op1, char op2) {
return (op1 != '*' && op1 != '/') || (op2 != '+' && op2 != '-');
}
public double evaluate(String input) {
Stack<Double> values = new Stack<>();
Stack<Character> ops = new Stack<>();
int stringIndex = 0;
while (stringIndex < input.length()) {
StringBuilder multiDigitsNumber = new StringBuilder();
// If the input is a number, put it onto the values stack
if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9') {
while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9') {
multiDigitsNumber.append(input.charAt(stringIndex++));
}
values.push(Double.parseDouble(multiDigitsNumber.toString()));
}
// If the input is an operator, put it onto the ops stack
else {
while (!ops.empty() && hasPrecedence(input.charAt(stringIndex), ops.peek())) {
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
}
ops.push(input.charAt(stringIndex++));
}
}
// Execute remaining operators in the values stack
while (!ops.empty()) {
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
}
// The final number in the values stack is the result
return values.pop();
}
Input example:
> 12+24*2-30/5.....
上述代码正常工作，但我想知道是否有方法可以替代下面这部分：
while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9') {
multiDigitsNumber.append(input.charAt(stringIndex++));
}
以避免在这种情况下使用嵌套的 while 循环。目标是在字符串中捕获数字，直到遇到运算符。提前感谢您的帮助。
英文:
I tried to Evaluate Mathematical Expressions in Java with the following code:
public double applyOp(char op,double b,double a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
return a / b;
}
return 0;
}
public boolean hasPrecedence(char op1,char op2)
{
return (op1 != '*' && op1 != '/') || (op2 != '+' && op2 != '-');
}
public double evaluate(String input) {
Stack<Double> values = new Stack<>();
Stack<Character> ops = new Stack<>();
int stringIndex = 0;
while (stringIndex < input.length())
{
StringBuilder multiDigitsNumber = new StringBuilder();
// If the input is number put to stack values
if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
multiDigitsNumber.append(input.charAt(stringIndex++));
}
values.push(Double.parseDouble(multiDigitsNumber.toString()));
}
// If the input is operator put to stack ops
else
{
while (!ops.empty() && hasPrecedence(input.charAt(stringIndex),ops.peek()))
{
values.push(applyOp(ops.pop(),values.pop(),values.pop()));
}
ops.push(input.charAt(stringIndex++));
}
}
// Execute remain operator in stack values
while (!ops.empty()) {
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
}
// The final number in stack value is result
return values.pop();
}
Input example:
> 12+24*2-30/5.....
The code above works fine but I wonder are there any way to replace
while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
multiDigitsNumber.append(input.charAt(stringIndex++));
}
with something else so I don't have to use nested while loop in this situation. The goal is to catch number in string until it reach an operator
Thanks in advance
答案1
得分: 1
可以像这样使用正则表达式（Regex）。
if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
String number = input.substring(stringIndex).replaceAll("^(\\d+).*", "$1");
values.push(Double.parseDouble(number));
stringIndex += number.length();
}
英文:
You can use Regex like this.
if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
String number = input.substring(stringIndex).replaceAll("^(\\d+).*", "$1");
values.push(Double.parseDouble(number));
stringIndex += number.length();
}
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