java - java中如何查找二进制字符串的子字符串

Question

String s="101010101010";
String sub=""; //substring
int k=2;


   package coreJava;
    import java.util.Scanner; 
    public class substring {    
           public static void main(String args[])
           {
              String string, sub;
              int k, c, i;

              Scanner in = new Scanner(System.in);
              System.out.println("Enter a string to print it's all substrings");
              string  = in.nextLine();

              i = string.length();   

              System.out.println("Substrings of \""+string+"\" are :-");

              for( c = 0 ; c < i ; c++ )
              {
                 for( k = 1 ; k <= i - c ; k++ )
                 {
                    sub = string.substring(c, c+k);
                    System.out.println(sub);
                 }
              }
           }
        }

1. 取一个二进制字符串 s="1010011010";//等等
2. 取一个变量k=2；
3. 取另一个变量i；//这是子串的长度(i>k)

现在我想找到上面字符串的子字符串，这样如果k=2，子字符串中1的数量必须是2，如果k=3，子字符串中1的数量必须是3，依此类推...

Output should be like this: 
string s="1010011010" 
Enter value of k=2; 
Enter length of substring i=3; 
substring= 101 110 101 011

Answer 1

创建一个“窗口”，其长度为您沿字符串移动的所需子字符串的长度，并维护当前窗口中 1 的数量。每次迭代时，您都会沿着一个窗口移动窗口，测试当前窗口之外的下一个字符、当前窗口中的第一个字符并相应地更新计数。在每次迭代期间，如果您的计数等于所需的长度，则打印当前窗口中的子字符串。

public class Substring {

    public static void main(String[] args) {
        String str = "1010011010";

        int k = 2;
        int i = 3;

        printAllSubstrings(str, i, k);

    }

    private static void printAllSubstrings(String str, int substringLength, int numberOfOnes) {
        // start index of the current window
        int startIndex = 0;

        // count of 1s in current window
        int count = 0;

        // count 1s in the first i characters
        for (int a = 0; a < substringLength; a++) {
            if (str.charAt(a) == '1') {
                count++;
            }
        }

        while (startIndex < str.length() - substringLength + 1) {
            if (count == numberOfOnes) {
                System.out.print(str.substring(startIndex, startIndex + substringLength));
                System.out.print(" ");
            }
            // Test next bit, which will be inside the window next iteration
            if (str.length() > startIndex + substringLength && str.charAt(startIndex + substringLength) == '1') {
                count ++;
            }
            // Test the starting bit, which will be outside the window next iteration
            if (str.charAt(startIndex) == '1') {
                count --;
            }
            startIndex++;
        }   
    }
}

输出:

101 011 110 101