我想检查一个数字是否连续，例如：
12345是连续的，但54321也是连续的。
现在我使用的正则表达式只说12345是连续的。
String consecutive = "^(?:0?123(45?)?|1?23456?|2?34567?|3?45678?|4?56789?|(5?6)?7890?|\n" +
" (0?1)?2345678?(90$)?|1?23456789?|2?345678(90?)?)$";
如何更改此正则表达式以匹配另一个？它必须是5位数。
这个正则表达式有更好的方法吗？
答案 0 :(得分：3)
我不认为正则表达式是这个问题的正确答案。尽管如此（但不寒而栗！）：
Pattern regex = Pattern.compile(
"\\b \n" +
"(?: \n" +
" (?: \n" +
" 0(?=1|\\b)| \n" +
" 1(?=2|\\b)| \n" +
" 2(?=3|\\b)| \n" +
" 3(?=4|\\b)| \n" +
" 4(?=5|\\b)| \n" +
" 5(?=6|\\b)| \n" +
" 6(?=7|\\b)| \n" +
" 7(?=8|\\b)| \n" +
" 8(?=9|\\b)| \n" +
" 9\\b # or 9(?=0|\\b) if you want to allow 890123\n" +
" )+ \n" +
" | \n" +
" (?: \n" +
" 9(?=8|\\b)| \n" +
" 8(?=7|\\b)| \n" +
" 7(?=6|\\b)| \n" +
" 6(?=5|\\b)| \n" +
" 5(?=4|\\b)| \n" +
" 4(?=3|\\b)| \n" +
" 3(?=2|\\b)| \n" +
" 2(?=1|\\b)| \n" +
" 1(?=0|\\b)| \n" +
" 0\\b # or 0(?=9|\\b) if you want to allow 321098\n" +
" )+ \n" +
") \n" +
"\\b",
Pattern.COMMENTS);
Matcher regexMatcher = regex.matcher(subjectString);
答案 1 :(得分：1)
正如这个问题的许多其他评论员，我认为这不应该用正则表达式完成。如果你想匹配重复的数字模式（如111），而不是任意数字序列（如123或246），则正则表达式很酷。由于我对这个主题感兴趣，我编写了这段代码。
输入字符串可以是任何字符序列（与非数字混合的数字），并且将返回给定长度的所有连续数字。
public class ConNumberParser {
public static List<String> findConsecutiveNumbers(String s, int length) {
LinkedList<String> matches = new LinkedList<>();
// seek to first digit
int i = 0;
while (i < s.length() && !Character.isDigit(s.charAt(i))) {
i++;
}
// store the beginning of a consecutive series
Integer matchedIndex = i;
// and the direction
Boolean increasing = null;
for (i++; i < s.length(); i++) {
final char c = s.charAt(i);
if (Character.isDigit(c)) {
if (null == matchedIndex) {
// first digit after other characters
matchedIndex = i;
increasing = null;
continue;
}
int difference = Character.getNumericValue(c) - Character.getNumericValue(s.charAt(i - 1));
if (Math.abs(difference) > 1) {
// no conescutive digits
matchedIndex = i;
increasing = null;
continue;
}
if (length > 2) {
if (null == increasing) {
// found first consecutive digit
increasing = (difference == 1);
continue;
}
final int expectedDiff = increasing ? 1 : -1;
if (difference != expectedDiff) {
// no conescutive digits in the right direction
matchedIndex = i - 1;
increasing = (difference == 1);
continue;
}
}
if (i - matchedIndex + 1 == length) {
// consecutive digits of given length found
matches.add(s.substring(matchedIndex, matchedIndex + length));
matchedIndex++; // move by one to keep matching overlapping
// sequences
}
} else {
matchedIndex = null;
}
}
return matches;
}
public static void main(String[] args) {
String test = "A12345b321";
List<String> m = findConsecutiveNumbers(test, 3);
System.out.print(test + " length=3: ");
System.out.println(m.toString());
String test2 = "Ax.*1290134543b3210";
List<String> m2 = findConsecutiveNumbers(test2, 3);
System.out.print(test2 + " length=3: ");
System.out.println(m2.toString());
List<String> m3 = findConsecutiveNumbers(test2, 2);
System.out.print(test2 + " length=2: ");
System.out.println(m3.toString());
}
}
产生预期结果：
A12345b321 length=3: [123, 234, 345, 321]
Ax.*1290134543b3210 length=3: [345, 543, 321, 210]
Ax.*1290134543b3210 length=2: [12, 01, 34, 45, 54, 43, 32, 21, 10]
答案 2 :(得分：0)
发现regex很难，所以我这样做了：
public static boolean isConsecutive(final String pinCode)
{
int [] digits = new int [pinCode.length()];
int [] differences = new int [pinCode.length()-1];
int temp = 0;
for(int i = 0; i < pinCode.length(); i++)
digits[i] = Integer.parseInt(String.valueOf(pinCode.charAt(i)));
for(int i = 0; i < digits.length -1; i++)
differences[i] = Math.abs(digits[i] - digits[i+1]);
if(differences.length != 0) {
temp = differences[0];
for (int i = 1; i < differences.length; i++)
if (temp != differences[i])
return false;
}
return true;
}
答案 3 :(得分：0)
这个怎么样？
function test(r) {
s = r.toString();
if(s.length == 2) {
return Math.abs(parseInt(s.split('')[0]) - parseInt(s.split('')[s.length-1])) == 1;
} else {
t = parseInt(s.split('')[0]) + parseInt(s.split('')[s.length-1]);
s = '' + (parseInt(s) + parseInt(s.split('').reverse().join(''))) / t;
return new RegExp('1{' + s.split('').length + '}').test(s);
}
}
答案 4 :(得分：-3)
它需要是正则表达式吗？在我的头顶，您可以将任何包含的字符串拆分为字符，转换为数字，然后检查b = a + 1 || b = a - 1（在开始时设置一些标志，以确定是否正在递增或递减。
修改
帖子here可能会帮助您。
这是它建议的正则表达式：
/(^|(.)(?!\2))(\d)\3{3}(?!\3)/（N.B.这仅适用于4个字符）