public class Model {
private Scanner sc = new Scanner(System.in);
private String expression;
private String exp[];
public Model()
{
expression = sc.nextLine();
split();
}
public void split()
{
//splitting the entered expression to array of operators alone and array of the numbers then create Arraylist to combine the operators and numbers together as if it is a string expression but as an array
String num[]= this.expression.split("[/+/*/-]");
String preop[]= this.expression.split("[0123456789]"); // this will give [empty, operator, operator...] therefore we will create another array to fill in the ops excluding empty
System.out.println("Test: Printing num Array");
for(int i = 0; i<num.length;i++)
{
System.out.print(num[i]+",");
}
System.out.println("\nTest: Printing preOp Array");
for(int i = 0; i<preop.length;i++)
{
System.out.print(preop[i]+ ",");
}
ArrayList<String> op = new ArrayList<>();//I used arraylist because easier
for(int i = 1; i<preop.length;i++)
{
op.add(preop[i]);
}
System.out.println("\nTest of the fixed preOp array: " + op);
//putting the operands and the operators together in the same array
ArrayList<String> exp = new ArrayList<>();
//fill the arraylist with numbers then add the operators to it by using number (index of the operator +1 +count)
for(int i = 0; i <num.length;i++)
{
exp.add(num[i]);
}
int count = 0;
for(int i = 0; i <op.size();i++)
{
exp.add(i+1+count, op.get(i));
count++;
}
System.out.println("Test: " + exp);
}
问题是,每当用户在表达式中输入两位数时,op数组就会给出空槽[op,op,empty,op]。
当用户输入一位数字时,我预期会出现类似的结果,如图像input with one digit numbers所示
2条答案
按热度按时间v1l68za41#
就是因为这个
我们用一个数字来分割,所以43也被分割成两部分,中间有一个空字符串。另外,你不需要命名正则表达式中的所有数字,你可以只做一个范围
"[0-9]"
。如果你想匹配一个或多个数字,添加一个+
。这应该可以工作:68bkxrlz2#
你可以使用正则表达式在任何一个操作符处拆分,然后使用lookahead和lookbehind来保持它们,从而一次性地解决这个问题:
输出: