I thought a problem for a day but still cannot solve it.
I have a formula input like "11+1+1+2". without space
I want to split the formula according to the operator.
Then I wrote like these:
String s = "11+1+1+2";
String splitByOp[] = s.split("[+|-|*|/|%]");
for(int c=0; c < splitByOp.length; c++){
System.out.println(splitByOp[c]);
The output is:
11
1
1
2
I want to put the operand(the output) and also the operator(+) into an ArrayList. But how can I keep the operator after spliting them?
I try to have one more Array to split the number.
String operator[] = s.split("\\d");
But the result is 11 become 1 1. The length of operator[] is 5.
In other words, how can I perform like:
The output:
11
+
1
+
1
+
2
You need to split on a regex that is non consuming. Specifically, on "word boundary":
String[] terms = s.split("\\b");
A "word boundary" is the gap between the word char and a non-word char, but digits are classified as word chars. Importantly, the match is non-consuming, so all of the content of the input is preserved in the split terms.
Here's some test code:
String s = "11+1+1+2";
String[] terms = s.split("\\b");
for (String term : terms)
System.out.println(term);
Output:
11
+
1
+
1
+
2
public static void main(String[] args) {
String s = "11+1+1+2";
String[] terms = s.split("(?=[+])|(?<=[+])");
System.out.println(Arrays.toString(terms));
}
output
[11, +, 1, +, 1, +, 2]
You could combine lookahead/lookbehind assertions
String[] array = s.split("(?=[+])|(?<=[+])");
Related
I have to convert a binary number to a hex number. The way I have decided to do this is to split the binary string into several strings of length 4 and assign each string its corresponding value in hex number (i.e. 1000 = 8, 1101 = D).
I have seen several question asking for a way to split a string into strings of size 4 the same thing but all of those solutions used a regex that gave a single string. For example I found this line of code in a solution:
System.out.println(Arrays.toString("String".split("(?<=\G.{4})")));
When I tried to use it with the binary number "10011000", I got "[1001, 1000]" but as a single string (the brackets, comma, and blank space were included as characters) and I was left with the same problem, how do I split a string.
Is there a way to split a string into an array of smaller strings?
You can try making the string a char array and then into another array of strings, add each 4 characters of the char array.
Try this:
String BinaryNumber = "10011010";
char[] n = new char[BinaryNumber.length()];
for(int i=0; i<BinaryNumber.length(); i++){
n[i] = BinaryNumber.charAt(i);
}
String str;
String[] NumberArray = new String[(BinaryNumber.length())/4];
int count = 0;
for(int i=0; i<BinaryNumber.length(); i+=4){
str = String.valueOf(n[i])+String.valueOf(n[i+1])+String.valueOf(n[i+2])+String.valueOf(n[i+3]);
NumberArray[count] = str;
count++;
}
I think this might be the solution, though it will only work if the length of the BinaryNumber is divisible by 4.
Try it like this.
String binaryNumber = "110101111";
// first make certain the binary string is a multiple of length four so
// pad on the left with 0 bits.
binaryNumber = "0".repeat(3 - (binaryNumber.length()+3) % 4)
+ binaryNumber;
// Then you can just split it like this as you described.
String[] groups = binaryNumber.split("(?<=\\G.{4})");
for (String v : groups) {
System.out.println(v);
}
prints
0001
1010
1111
This question is similar to my previous question Split a string contain dash and minus sign. But I asked it in a wrong and then it got a slightly different semantics and people answered(including) in that perspective. Therefore rather than modifying that question I thought it's better to ask in a new question.
I have to split a string which contain hyphen-minus character and minus sign. I tried to split based on the unicode character (https://en.wikipedia.org/wiki/Hyphen#Unicode), still it considering minus sign same as hyphen-minus character. Is there a way I can solve it?
Expected output
(coun)
(US)
-1
Actual output
(coun)
(US)
// actually blank line will print here but SO editor squeezing the blank line
1
public static void main(String[] args) {
char dash = '-';
int i = -1;
String a = "(country)" + dash + "(US)" + dash + i;
Pattern p = Pattern.compile("-", Pattern.LITERAL);
String[] m = p.split(a);
for (String s : m) {
System.out.println(s);
}
}
char dash = '\u2010'; // 2010 is hyphen, 002D is hyphen-minus
int i = -1;
String a = "(country)" + dash + "(US)" + dash + i;
Pattern p = Pattern.compile("\u2010", Pattern.LITERAL);
String[] m = p.split(a);
for (String s : m) {
System.out.println(s);
}
The string representation of an integer always uses the hyphen-minus as the negative sign:
From Integer.toString:
If the first argument is negative, the first element of the result is the ASCII minus character '-' ('\u002D'). If the first argument is not negative, no sign character appears in the result.
so in the end your string has 3 hyphen-minus characters. That's why split can't distinguish between them.
Since you can't change the string representation of an integer, you need to change the dash variable to store a hyphen instead of hyphen-minus. Now there are 2 hyphens and 1 hyphen-minus in your string, making split able to distinguish between them.
This question already has answers here:
How to split a string with any whitespace chars as delimiters
(13 answers)
Closed 5 years ago.
I'm looking for a way to convert a string to an array and strip all whitespaces in the process. Here's what I have:
String[] splitArray = input.split(" ").trim();
But I can't figure out how to get rid of spaces in between the elements.
For example,
input = " 1 2 3 4 5 "
I want splitArray to be:
[1,2,3,4,5]
First off, this input.split(" ").trim(); won't compile since you can't call trim() on an array, but fortunately you don't need to. Your problem is that your regex, " " is treating each space as a split target, and with an input String like so:
String input = " 1 2 3 4 5 ";
You end up creating an array filled with several empty "" String items.
So this code:
String input = " 1 2 3 4 5 ";
// String[] splitArray = input.split("\\s+").trim();
String[] splitArray = input.trim().split(" ");
System.out.println(Arrays.toString(splitArray));
will result in this output:
[1, , , , , , , , 2, 3, 4, , , , , , 5]
What you need to do is to create a regex that greedily groups all the spaces or whitespace characters together, and fortunately we have this ability -- the + operator
Simply use a greedy split with the whitespace regex group
String[] splitArray = input.trim().split("\\s+");
\\s denotes any white-space character, and the trailing + will greedily aggregate one or more contiguous white-space characters together.
And actually, in your situation where the whitespace is nothing but multiples of spaces: " ", this is adequate:
String[] splitArray = input.trim().split(" +");
Appropriate tutorials for this:
short-hand character classes -- discusses \\s
repetition -- discusses the + also ? and * repetition characters
Try:
String[] result = input.split(" ");
I have a function which takes a String containing a math expression such as 6+9*8 or 4+9 and it evaluates them from left to right (without normal order of operation rules).
I've been stuck with this problem for the past couple of hours and have finally found the culprit BUT I have no idea why it is doing what it does. When I split the string through regex (.split("\\d") and .split("\\D")), I make it go into 2 arrays, one is a int[] where it contains the numbers involved in the expression and a String[] where it contains the operations.
What I've realized is that when I do the following:
String question = "5+9*8";
String[] mathOperations = question.split("\\d");
for(int i = 0; i < mathOperations.length; i++) {
System.out.println("Math Operation at " + i + " is " + mathOperations[i]);
}
it does not put the first operation sign in index 0, rather it puts it in index 1. Why is this?
This is the system.out on the console:
Math Operation at 0 is
Math Operation at 1 is +
Math Operation at 2 is *
Because on position 0 of mathOperations there's an empty String. In other words
mathOperations = {"", "+", "*"};
According to split documentation
The array returned by this method contains each substring of this
string that is terminated by another substring that matches the given
expression or is terminated by the end of the string. ...
Why isn't there an empty string at the end of the array too?
Trailing empty strings are therefore not included in the resulting
array.
More detailed explanation - your regex matched the String like this:
"(5)+(9)*(8)" -> "" + (5) + "+" + (9) + "*" + (8) + ""
but the trailing empty string is discarded as specified by the documentation.
(hope this silly illustration helps)
Also a thing worth noting, the regex you used "\\d", would split following string "55+5" into
["", "", "+"]
That's because you match only a single character, you should probably use "\\d+"
You may find the following variation on your program helpful, as one split does the jobs of both of yours...
public class zw {
public static void main(String[] args) {
String question = "85+9*8-900+77";
String[] bits = question.split("\\b");
for (int i = 0; i < bits.length; ++i) System.out.println("[" + bits[i] + "]");
}
}
and its output:
[]
[85]
[+]
[9]
[*]
[8]
[-]
[900]
[+]
[77]
In this program, I used \b as a "zero-width boundary" to do the splitting. No characters were harmed during the split, they all went into the array.
More info here: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
and here: http://www.regular-expressions.info/wordboundaries.html
Life is very easy if the expression has values from 0 to 9 but
If expression = 23+52*5 is input by user then I take it in a String named expression.
Now what I want is that a new String or char Array in such a fashion that:
String s or char[] ch = ['23','+','52','*','5']
so that ch[0] or s.charAt(0) gives me 23 and not 2.
To do so I have tried the following and am stuck on what to do next:
for(int i=0;i<expression.length();i++)
{
int operand = 0;
while(i<expression.length() && sol.isOperand(expression.charAt(i))) {
// For a number with more than one digits, as we are scanning
// from left to right.
// Everytime , we get a digit towards right, we can
// multiply current total in operand by 10
// and add the new digit.
operand = (operand*10) + (expression.charAt(i) - '0');
i++;
}
// Finally, you will come out of while loop with i set to a non-numeric
// character or end of string decrement it because it will be
// incremented in increment section of loop once again.
// We do not want to skip the non-numeric character by
// incrementing it twice.
i--;
/**
* I have the desired integer value but how to put it on a string
* or char array to fulfill my needs as stated above.
**/
// help or ideas for code here to get desired output
}
In the while loop the method isOperand(char) is returning a boolean value true if char provided is >=0 or <=9.
You're going to want a String[] when you break the expression apart. Regex lookbehind/lookahead (Java Pattern Reference) allows you to split a String and keep the delimiters. In this case, your delimiters are the operands. You can use a split pattern like this:
public static void main(String[] args) throws Exception {
String expression = "23+52*5";
String[] pieces = expression.split("(?<=\\+|-|\\*|/)|(?=\\+|-|\\*|/)");
System.out.println(Arrays.toString(pieces));
}
Results:
[23, +, 52, *, 5]
One this to consider, is if your expression contains any spaces then you're going to want to remove them before splitting the expression apart. Otherwise, the spaces will be included in the results. To remove the spaces from the expression can be done with String.replaceAll() like this:
expression = expression.replaceAll("\\s+", "");
The "\\s+" is a regular expression pattern that means a whitespace character: [ \t\n\x0B\f\r]. This statement replaces all whitespace characters with an empty space essentially removing them.
String expr = "23+52*5";
String[] operands = expr.split("[^0-9]");
String[] operators = expr.split("[0-9]+");
This breaks it into:
try using a switch case:
//read first character
switch data
if number : what is it ( 0,1,2,3...?)
save the number
if its an operator
save operator
** 2 the switch has 4 cases for operators and 10 for digits **
// read the next character
again switch
if its a number after a number add both chars into a string
if its an operator "close" the number string and convert it into an int
If you will need some code I will gladly help.
Hope the answer is useful.
for i = 1 ; i = string.length
switch :
case 1,2,3,4,5,6,7,8,9,0 (if its a digit)
x(char) = char(i)
case +,-,*,/ (operator)
y(char) = char(i)
x[1] + x[2] +...x[n] = X(string) // build all the digits saves before the operator into 1 string
convert_to_int(X(string))
You can use regex:
\d+
Will grab any instance of one or more consecutive digits.