如何在整数数组中找到重复的整数序列？

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如何在整数数组中找到重复的整数序列？

How to find repeating sequence of Integers in an array of Integers?

00将重复，123123也是如此，但01234593623不会重复。

00 would be repeating, so would 123123, but 01234593623 would not be.

我知道如何做到这一点，但在我看来它很模糊，而且由于这个原因，我的实施并不顺利。

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.

我的想法是

每次进行for循环时抵消一定金额

在其中循环并比较那个偏移的数字块

在Java中，我得到了这个：

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"); }

使用这些参数运行时：

repeatingSeqOf(2, new String[] {"1", "2", "3", "4", "5", "6", "7", "7" });

我正在填充节数组，但它在索引超出范围异常时中断。

I am correctly filling the section arrays, but it breaks on an index out of bounds exception.

推荐答案

@MiljenMikic答案很棒，特别是因为语法实际上不是常规的。 ：D

@MiljenMikic answer's is great, especially since the grammar isn't actually regular. :D

如果你想在一般的数组上做这个，或者想要理解它，这几乎就是正则表达式所做的：

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); } } } }