问题
How to find repeating sequence of Integers in an array of Integers?
00 would be repeating, so would 123123, but 01234593623 would not be.
I have an idea to how to do this, but it is blurry in my mind, and my implementation doesn't go far due to this.
My idea was
- Offset by a certain amount each time going through a for loop
- Loop through inside of that and compare chunks of numbers by that offset
In Java, I got this far:
String[] p1 = new String[nDigitGroup];
String[] p2 = new String[nDigitGroup];
for (int pos = 0; pos < number.length - 1; pos++)
{
System.out.println("HERE: " + pos + (nDigitGroup - 1));
int arrayCounter = -1;
for (int n = pos; n < pos + nDigitGroup ; n++)
{
System.out.printf("\nPOS: %d\nN: %d\n", pos, n);
arrayCounter++;
p1[arrayCounter] = number[n];
System.out.println(p1[arrayCounter]);
}
pos += nDigitGroup;
arrayCounter = -1;
System.out.println("SWITCHING");
for (int n = pos; n < pos + nDigitGroup ; n++)
{
System.out.printf("\nPOS: %d\nN: %d\n", pos, n);
arrayCounter++;
p2[arrayCounter] = number[n];
System.out.println(p2[arrayCounter]);
}
if (p1[0].equals(p2[0]) && p1[1].equals(p2[1])) System.out.println("MATCHING");
}
When ran with these arguments:
repeatingSeqOf(2, new String[] {"1", "2", "3", "4", "5", "6", "7", "7" });
I am correctly filling the section arrays, but it breaks on an index out of bounds exception.
回答1:
@MiljenMikic answer's is great, especially since the grammar isn't actually regular. :D
If you want to do it on an array in general, or want to understand it, this does pretty much exactly what the regex does:
public static void main(String[] args) {
int[] arr = {0, 1, 2, 3, 2, 3}; // 2, 3 repeats at position 2.
// for every position in the array:
for (int startPos = 0; startPos < arr.length; startPos++) {
// check if there is a repeating sequence here:
// check every sequence length which is lower or equal to half the
// remaining array length: (this is important, otherwise we'll go out of bounds)
for (int sequenceLength = 1; sequenceLength <= (arr.length - startPos) / 2; sequenceLength++) {
// check if the sequences of length sequenceLength which start
// at startPos and (startPos + sequenceLength (the one
// immediately following it)) are equal:
boolean sequencesAreEqual = true;
for (int i = 0; i < sequenceLength; i++) {
if (arr[startPos + i] != arr[startPos + sequenceLength + i]) {
sequencesAreEqual = false;
break;
}
}
if (sequencesAreEqual) {
System.out.println("Found repeating sequence at pos " + startPos);
}
}
}
}
回答2:
You can always play with regular expressions to achieve a desired result. Use the regex backreference and combine it with the greedy quantifier:
void printRepeating(String arrayOfInt)
{
String regex = "(\\d+)\\1";
Pattern patt = Pattern.compile(regex);
Matcher matcher = patt.matcher(arrayOfInt);
while (matcher.find())
{
System.out.println("Repeated substring: " + matcher.group(1));
}
}
回答3:
The answer posted by @AdrianLeonhard is perfectly working. But if I have a sequence of 0, 1, 2, 3, 4, 3, 5, 6, 4, 7, 8, 7, 8 many might be wondering on how to get all the repeated numbers from the array.
So, I wrote this simple logic which prints all the repeated numbers with their positions
int[] arr = {0, 1, 2, 3, 4, 3, 5, 6, 4, 7, 8, 7, 8};
for(int i=0; i<arr.length;i++){
for(int j=i+1; j<arr.length;j++){
if(arr[i] == arr[j]){
System.out.println("Number: "+arr[i]+" found repeating at position: "+i+" , repeated at position "+j);
}
}
}
回答4:
Try this:
string lookIn = "99123998877665544123";
// above has length of 20 (in positions 0 through 19)
int patternLength = 3;
// want to search each triple of letters 0-2, 1-3, 2-4 ... 17-19
// however since there must be 3 chars after the 3-char pattern
// we only want to search the triples up to 14-16 (20 - 3*2)
for (int i=0; i <= lookIn.Length - patternLength * 2; i++) {
string lookingFor = lookIn.Substring(i, patternLength);
// start looking at the pos after the pattern
int iFoundPos = lookIn.IndexOf(lookingFor, i + patternLength);
if (iFoundPos > -1) {
string msg = "Found pattern '" + lookingFor
+ "' at position " + i
+ " recurs at position " + iFoundPos;
}
}
// of course, you will want to validate that patternLength is less than
// or equal to half the length of lookIn.Length, etc.
EDIT: improved and converted to javascript (from C# ... oops, sorry about that...)
function testfn() {
var lookIn = "99123998877665544123";
// above has length of 20 (in positions 0 through 19)
var patternLength_Min = 2;
var patternLength_Max = 5;
if (patternLength_Max > (lookIn.length / 2)
|| patternLength_Max < patternLength_Min
|| patternLength_Min < 1) {
alert('Invalid lengths.')
}
var msg = "";
for (var pLen = patternLength_Min; pLen <= patternLength_Max; pLen++) {
for (var i = 0; i <= lookIn.length - pLen * 2; i++) {
var lookingFor = lookIn.substring(i, i + pLen);
// start looking at the pos after the pattern
var iFoundPos = lookIn.indexOf(lookingFor, i + pLen);
if (iFoundPos > -1) {
msg = msg + "Found '" + lookingFor
+ "' at pos=" + i
+ " recurs at pos=" + iFoundPos + "\n";
;
}
}
}
alert(msg);
}
The message box displays the following:
Found '99' at pos=0 recurs at pos=5
Found '12' at pos=2 recurs at pos=17
Found '23' at pos=3 recurs at pos=18
Found '123' at pos=2 recurs at pos=17
来源：https://stackoverflow.com/questions/28847164/how-to-find-repeating-sequence-of-integers-in-an-array-of-integers