英文:
How to determine if a number is binary and has an even amount of zeros with only one regex?
问题
我想要一个正则表达式来判断一个数字是否为二进制，并且其中的零数量为偶数。
例如，如果我们考虑这三个数字：
- 10001010
- 00010010
- 45671892
我希望正则表达式匹配以下内容：
- 10001010 // 由于零的数量不均匀，所以不匹配
- 00010010 // 因为零的数量为偶数，所以匹配
- 45671892 // 不匹配，因为它不是二进制数
目前的解决方案如下：
[1]*(0[1]*0[1]*)*
我已经在 https://regex101.com/ 上进行了测试，对我来说它是有效的。
问题仍然在线上，只是为了看是否有人能提出更有效的解决方案。
非常感谢每一次的帮助。
提前谢谢！
英文:
I want a regex to determine if a number is binary and if it has an even amount of zeros in it.
For example, if we consider these 3 numbers:
- 10001010
- 00010010
- 45671892
I want the regex to match the following:
- 10001010 //Does NOT match because it has an uneven amount of zeros
- 00010010 //Does match because it has an even amount of zeros
- 45671892 //Does not match because it is not a binary number
The current solution is the following:
[1]*(0[1]*0[1]*)*
I tested it with https://regex101.com/ and it works for me.
The question is still online just to see if anyone can come up with a more efficient solution
I am thankful for every help.
Thank you in advance!
答案1
得分: 3
^[1]*(0[1]*0[1]*)*$
这个正则表达式可以这样解释：
查找任意数量的`1`，后面跟随一对`00`，中间可以有零个或多个`1`。
英文:
^[1]*(0[1]*0[1]*)*$
which can be interpreted in the following way:
find any number of 1 followed by pairs of 00 with optional one or more 1s in between.
答案2
得分: 2
你可以使用 Long#parseLong 方法来检查给定的数字字符串是否表示二进制数字。如果是的话，你可以从数字字符串中获得一个流，筛选出 0，计算 0 的数量，并检查数量是否能被 2 整除，从而确定有奇数个还是偶数个 0。
public class Main {
public static void main(String[] args) {
String[] arr = { "10001010", "00010010", "45671892" };
for (String s : arr) {
// 检查数字字符串是否为二进制
try {
// 可以验证最多 64 位
Long.parseLong(s, 2);
if (s.chars().filter(c -> c == '0').count() % 2 == 0) {
System.out.println(s + " 表示二进制数字且有偶数个 0。");
} else {
System.out.println(s + " 表示二进制数字且有奇数个 0。");
}
} catch (Exception e) {
System.out.println(s + " 不表示二进制数字。");
}
}
}
}
输出结果:
10001010 表示二进制数字且有奇数个 0。
00010010 表示二进制数字且有偶数个 0。
45671892 不表示二进制数字。
英文:
You can use Long#parseLong to check if the given number string represents a binary number or not. If yes, you can get a stream out of the number string, filter for 0, count the number of 0s and check if the count is divisible by 2 or not to determine if it has odd or even number of 0s.
public class Main {
public static void main(String[] args) {
String[] arr = { "10001010", "00010010", "45671892" };
for (String s : arr) {
// Check if the number string is binary
try {
// Can validate up to 64 bits
Long.parseLong(s, 2);
if (s.chars().filter(c -> c == '0').count() % 2 == 0) {
System.out.println(s + " represents a binary number and has an even number of 0s.");
} else {
System.out.println(s + " represents a binary number and has an odd number of 0s.");
}
} catch (Exception e) {
System.out.println(s + " does not represent a binary number.");
}
}
}
}
Output:
10001010 represents a binary number and has an odd number of 0s.
00010010 represents a binary number and has an even number of 0s.
45671892 does not represent a binary number.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论