编码蝙蝠挑战：maxBlock

Question

给定CodingBat中的任务maxBlock：

给定一个字符串，返回字符串中最大“块”的长度。块是相同的相邻字符的运行。

maxBlock("hoopla") → 2
maxBlock("abbCCCddBBBxx") → 3
maxBlock("") → 0

我的解决方案通过了所有测试，除了一个：

public int maxBlock(String str) {
  int maxBlock = 0;
  int currentBlock = 1;
  
  if (str.length() < 1) {
    return maxBlock;
  } else {
    maxBlock = 1;
  }
  
  for (int i = 0; i < str.length() - 1; i++) {
    if (str.charAt(i) ==  str.charAt(i+1)) {
      currentBlock++;
      if (currentBlock > maxBlock) {
        maxBlock = currentBlock;
        currentBlock = 0;
      }
    }
  }
  
  return maxBlock;
}

Answer 1

此代码由IntelliJ的GitHub Copilot插件生成，您只需输入int maxBlock，然后GitHub Copilot生成其余代码：

private int maxBlock(String s) {
    if (s == null || s.length() == 0) return 0;
    int maxBlocks = 0;
    int currentBlocks = 0;
    char currentChar = s.charAt(0);
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == currentChar) {
            currentBlocks++;
        } else {
            currentBlocks = 1;
            currentChar = s.charAt(i);
        }
        maxBlocks = Math.max(maxBlocks, currentBlocks);
    }
    return maxBlocks;
}

我只添加了<代码>如果（s==null | | s.length（）=0）返回0 到生成的代码。

Answer 2

虽然我认为@Robby Cornelissen和@Eskandar Abedini的解决方案肯定比我的更可行，更接近原始代码，但我想通过使用正则表达式和一些编码逻辑来提供一种不同的方法。

下面的正则表达式查找字符重新出现0次或更多次的匹配项。这可以与Matcher类的results（）方法一起使用，以提取找到的最长匹配。

下面是测试regex的链接

https://regex101.com/r/OcMoQ9/1

List<String> testCase = List.of("hoopla", "abbCCCddBBBxx", "", "xyz", "xxyz", "xyzz", "abbbcbbbxbbbx", "XXBBBbbxx", "XXBBBbbxxXXXX", "XX2222BBBbbXX2222");
Pattern pattern = Pattern.compile("(.)(\\1*)");
Matcher matcher;

for (String s : testCase) {
    matcher = pattern.matcher(s);
    System.out.printf("%s => %d%n", s, matcher
            .results()
            .map(match -> match.group().length())
            .max(Comparator.comparing(Integer::intValue))
            .orElse(0));
}

//Or even simpliefied like so as @Holger has pointed out in the comments
for (String s : testCase) {
    matcher = pattern.matcher(s);
    System.out.printf("%s => %d%n", s, matcher
            .results()
            .mapToInt(match -> match.group().length())
            .max()
            .orElse(0));
}

https://www.jdoodle.com/iembed/v0/rH7

hoopla => 2
abbCCCddBBBxx => 3
 => 0
xyz => 1
xxyz => 2
xyzz => 2
abbbcbbbxbbbx => 3
XXBBBbbxx => 3
XXBBBbbxxXXXX => 4
XX2222BBBbbXX2222 => 4

Answer 3

您没有正确重置电流块计数。如果以下字符与当前字符不同，您需要将电流块重置为1。如果不是，您需要继续增加电流块：

public int maxBlock(String str) {
  int maxBlock = 0;
  int currentBlock = 1;
  
  if (str.length() < 1) {
    return maxBlock;
  } else {
    maxBlock = 1;
  }
  
  for (int i = 0; i < str.length() - 1; i++) {
    if (str.charAt(i) ==  str.charAt(i+1)) {
      currentBlock++;
      if (currentBlock > maxBlock) {
        maxBlock = currentBlock;
      }
    } else {
      currentBlock = 1;
    }
  }
  
  return maxBlock;
}