如何在整数数组中找到重复的整数序列？

如何在整数数组中找到重复的整数序列？

00将重复，123123也将重复，但01234593623不会。

我知道如何做到这一点，但在我看来它很模糊，而且我的实现并没有因此而走得太远。

我的想法是

1. 每次经过for循环时偏移一定量
2. 在其中循环，并通过该偏移比较数字块

在Java中，我得到了这个：

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" }); 

我正确填充节数组，但它打破了索引超出范围的exception。

Answer 1

@MiljenMikic答案很棒，特别是因为语法实际上并不常见。 ：d

如果你想在一般的数组上做这件事，或者想要理解它，这几乎就是正则表达式的作用：

 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); } } } } 

Answer 2

您始终可以使用正则表达式来实现所需的结果。 使用正则表达式反向引用并将其与贪婪量词结合使用：

  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)); } } 

Answer 3

@AdrianLeonhard发布的答案非常有效。 但是，如果我有一个0,1,2,3,4,3,5,6,4,7,8,7,8的序列，许多人可能想知道如何从arrays中获取所有重复的数字。

所以，我写了这个简单的逻辑，用他们的位置打印所有重复的数字

  int[] arr = {0, 1, 2, 3, 4, 3, 5, 6, 4, 7, 8, 7, 8}; for(int i=0; i 

Answer 4

尝试这个：

 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. 

编辑：改进并转换为javascript（来自C＃… oops，抱歉…）

 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); } 

消息框显示以下内容：

 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