英文:
How to split a string into several strings of equal sizes
问题
我需要将一个二进制数转换为十六进制数。我决定这样做的方法是将二进制字符串分成若干长度为4的子字符串，并为每个子字符串赋予其在十六进制数中对应的值（例如，1000 = 8，1101 = D）。
我看到有几个问题询问如何将字符串分成长度为4的子字符串，但所有这些解决方案都使用了一个正则表达式，导致得到一个单独的字符串。例如，我在一个解决方案中找到了这行代码：
System.out.println(Arrays.toString("String".split("(?<=\\G.{4})")));
当我尝试将其用于二进制数"10011000"时，我得到了"[1001, 1000]"，但作为一个单独的字符串（括号、逗号和空格都包含在字符中），我仍然面临同样的问题，如何将一个字符串分割。
有没有一种将字符串拆分为较小字符串数组的方法？
英文:
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?
答案1
得分: 1
你可以尝试将字符串转换为字符数组，然后再转换为另一个字符串数组，在字符数组的基础上每4个字符添加到新的字符串数组中。
尝试这样做：
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++;
}
我认为这可能是解决方案，但只有在BinaryNumber的长度能够被4整除时才有效。
英文:
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.
答案2
得分: 1
尝试这样做：
String binaryNumber = "110101111";
// 首先确保二进制字符串的长度是4的倍数，如果不是，在左边用0位进行填充。
binaryNumber = "0".repeat(3 - (binaryNumber.length()+3) % 4)
+ binaryNumber;
// 然后按照你所描述的方式分割字符串。
String[] groups = binaryNumber.split("(?<=\\G.{4})");
for (String v : groups) {
System.out.println(v);
}
输出结果为
0001
1010
1111
英文:
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
</details>
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