4

给定一个字符串：“blablafblafbla”和2个限制：x = 3，y = 5我想找到长度在x和y之间的最长重复子字符串。如果有很多，第一个在我的例子中是“ blaf" 几个问题： 1. 使用正则表达式更容易吗？2.我知道如何找到最长的子串，但是我必须在哪里设置它在 x 和 y 之间的条件？

public static String longestDuplicate(String text)
{
    String longest = "";
    for (int i = 0; i < text.length() - 2 * longest.length() * 2; i++)
    {
        OUTER: for (int j = longest.length() + 1; j * 2 < text.length() - i; j++)
        {
            String find = text.substring(i, i + j);
            for (int k = i + j; k <= text.length() - j; k++)
            {
                if (text.substring(k, k + j).equals(find))
                {
                    longest = find;
                    continue OUTER;
                }
            }
            break;
        }
    }
    return longest;
}
4

4 回答 4

2

您提供的代码是解决您遇到的问题的一种极其低效的方法。我将使用Rabin-Karp或其他一些滚动哈希算法来实现该解决方案，这将使您能够解决复杂性问题O((y-x) * L)。

您不能在这里使用正则表达式——它们旨在解决完全不同的任务。

至于您关于如何使用您的解决方案x来查找长度在and之间的最长子字符串的问题y，只需修改循环j以仅考虑区间内的值[x, y]。这是您如何做到这一点的方法。

for (int j = Math.max(longest.length() + 1, x) ; j * 2 < text.length() - i && j < y; j++)

编辑：要找到最长的子字符串，请反转 for 循环：

for (int j = Math.min((text.length() - i -1)/2, y) ; j > longest.length() && j >=x; j--)
于 2013-02-08T10:13:49.607 回答
1 
public static int commonPrefix (String string, int x, int y)
{
    int l = string.length ();
    int n = 0;
    int oy = y;
    while (x < oy && y < l && string.charAt (x) == string.charAt (y))
    {
        n++; x++; y++;
    }
    return n;
}

public static String longestRepeatingSubstring (
    String string, int minLength, int maxLength)
{
    String found = null; 

    int l = string.length ();
    int fl = minLength; 
    for (int x = 0; x < l - fl * 2; x++)
        for (int y = x + 1; y < l - fl; y++)
        {
            int n = commonPrefix(string, x, y);

            if (n >= maxLength)
                return string.substring(x, x + maxLength);

            if (n > fl)
            {
                found = string.substring (x, x + n);
                fl = n;
            }
        }

    return found;
}

public static void main(String[] args) {
    System.out.println (longestRepeatingSubstring ("blablafblafblafblaf", 3, 5));
}
于 2013-02-08T10:28:56.480 回答
1

这是一个使用正则表达式的笨拙实现：

//import java.util.regex.*;

public static String longestRepeatingSubstring (String string, int min, int max)
{
  for (int i=max; i>=min; i--){
    for (int j=0; j<string.length()-i+1; j++){

      String substr = string.substring(j,j+i);
      Pattern pattern = Pattern.compile(substr);
      Matcher matcher = pattern.matcher(string);

      int count = 0;
      while (matcher.find()) count++;

      if (count > 1) return substr;
    }
  }

  return null;
}

public static void main(String[] args) {
  System.out.println (longestRepeatingSubstring ("blablafblafbla", 3, 5));
}
于 2013-02-08T17:10:10.883 回答
0 
    public static int getCount(String string , String subString){

    int count = 0;
    int fromIndex = 0;
    do{
    if(string.indexOf(subString, fromIndex) != -1){
        count++;
        fromIndex = string.indexOf(subString, fromIndex);
    }
    }while(fromIndex == string.length()-1);
    return count;
}
public static String longestRepeatingSubstring (int min,int max , String string){
    Vector substrs = new Vector();
    Vector substrs_length = new Vector();
    for (int i=min; i<=max; i++){
        for (int j=0; j<string.length()-i+1; j++){
            String substr=string.substring(j, i+j);
            System.out.println(substr);
            if (substrs.indexOf(substr) == -1){
                int count =getCount(string, substr);
                if (count != 0) {
                    substrs.addElement(substr);
                    substrs_length.addElement(count);
                }
            }
        }
    }
    int maxLength = 0;
    int index = -1;
    for(int i = 0 ;  i < substrs_length.size() ; i++){
        int length = (int) substrs_length.elementAt(i);
        if(length > maxLength){
            maxLength = length;
            index = i;
        }
    }
    return (String) substrs.elementAt(index);
}
public static void main(String [] arg){
    System.out.print(longestRepeatingSubstring(3, 5, "blablafblafbla"));
}
于 2013-02-08T19:36:02.313 回答