无法确定如何继续：Java错误~

我正在重置不应该重置的临时块。有人能帮我格式化我的代码吗？我不知道如何放置代码标签。

我想你应该跟踪最长的1或0序列。如果是这种情况，那么尝试将检查maxBlock条件的两个If语句放入循环中。

我认为您的循环最好是这样：我删除了对“I”的一些冗余检查，并简化了对“maxBlock”的检查：

import java.util.Scanner;

public class Test {
  private static int decimalNum = 0;
  private static String binary = "";

  private static void getInput() {
    Scanner sc = new Scanner(System.in);
    System.out.println("Please type in a number");
    decimalNum = sc.nextInt();
  }

  private static void convert() {
    int decimalNumber = decimalNum;
    String binaryNumber;
    if (decimalNumber <= 0)
      System.out.println("ERROR: entered integer is nonpositive.");
    else {
      binaryNumber = "";
      while (decimalNumber != 0) {
        // add spaces to separate 4-digit groups
        if (binaryNumber.length() % 5 == 0)
          binaryNumber = "" + binaryNumber;
        // extract last digit in binary representation
        // and add it to binaryNumber
        binaryNumber = (decimalNumber % 2) + binaryNumber;
        // cut last digit in binary representation
        decimalNumber /= 2;
      }
      binary = binaryNumber;
      System.out.println("Binary: " + binaryNumber);
    }
  }

  public static void count() {
    String s = binary + "";
    System.out.println("Binary number: " + s);
    int temp1Block = 0;
    int temp0Block = 0;
    int maxBlock = 0;
    for (int i = 0; i < s.length(); i++) {
      if ((s.charAt(i) == '1') && (i < s.length())) {
        temp0Block = 0;
        temp1Block++;
      }
      if ((s.charAt(i) == '0') && (i < s.length())) {
        temp1Block = 0;
        temp0Block++;
      }
    }
    if (maxBlock < temp0Block) {
      maxBlock = temp0Block;
    }
    if (maxBlock < temp1Block) {
      maxBlock = temp1Block;
    }
    System.out.println("Maxblock " + maxBlock);
  }

  public static void main(String[] args) {
    getInput();
    convert();
    count();
  }
}

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只需点击SO编辑器中的“格式”按钮。您遇到了什么问题？您可能想先在IDE中格式化它-当单个方法的所有语句都在一行上时，很难理解！请不要将问题交叉发布到多个SE站点。堆栈溢出是回答您的问题的正确位置，尽管正如我在程序员身上所说的，它需要更多的细节。如果你在不该重置的时候重置了tempBlocks，那就停止吧？你的实际问题是什么？如果你告诉我们你想要什么，这会更容易回答！我需要重新设置。问题是我很难跟踪最大块数。我想粘贴答案，以防功能中有人：

for(int i = 0; i < s.length(); ++i) {
    if(s.charAt(i) == '1') {
        // in case we were just keeping track of 0's
        maxBlock = Math.max(maxBlock, temp0Count);
        temp0Count = 0;

        temp1Count++;
    }
    else {
        // in case we were just keeping track of 1's
        maxBlock = Math.max(maxBlock, temp1Count);
        temp1Count = 0;

        temp0Count++;
    }
}