寻找最长的重复子串

Question

解决这个问题的最佳方法（性能方面）是什么？ 我被推荐使用后缀树。这是最好的方法吗？

Answer 1

查看此链接：http://introcs.cs.princeton.edu/java/42sort/LRS.java.html

/*************************************************************************
 *  Compilation:  javac LRS.java
 *  Execution:    java LRS < file.txt
 *  Dependencies: StdIn.java
 *  
 *  Reads a text corpus from stdin, replaces all consecutive blocks of
 *  whitespace with a single space, and then computes the longest
 *  repeated substring in that corpus. Suffix sorts the corpus using
 *  the system sort, then finds the longest repeated substring among 
 *  consecutive suffixes in the sorted order.
 * 
 *  % java LRS < mobydick.txt
 *  ',- Such a funny, sporty, gamy, jesty, joky, hoky-poky lad, is the Ocean, oh! Th'
 * 
 *  % java LRS 
 *  aaaaaaaaa
 *  'aaaaaaaa'
 *
 *  % java LRS
 *  abcdefg
 *  ''
 *
 *************************************************************************/


import java.util.Arrays;

public class LRS {

    // return the longest common prefix of s and t
    public static String lcp(String s, String t) {
        int n = Math.min(s.length(), t.length());
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) != t.charAt(i))
                return s.substring(0, i);
        }
        return s.substring(0, n);
    }


    // return the longest repeated string in s
    public static String lrs(String s) {

        // form the N suffixes
        int N  = s.length();
        String[] suffixes = new String[N];
        for (int i = 0; i < N; i++) {
            suffixes[i] = s.substring(i, N);
        }

        // sort them
        Arrays.sort(suffixes);

        // find longest repeated substring by comparing adjacent sorted suffixes
        String lrs = "";
        for (int i = 0; i < N - 1; i++) {
            String x = lcp(suffixes[i], suffixes[i+1]);
            if (x.length() > lrs.length())
                lrs = x;
        }
        return lrs;
    }



    // read in text, replacing all consecutive whitespace with a single space
    // then compute longest repeated substring
    public static void main(String[] args) {
        String s = StdIn.readAll();
        s = s.replaceAll("\\s+", " ");
        StdOut.println("'" + lrs(s) + "'");
    }
}

Answer 2

也看看http://en.wikipedia.org/wiki/Suffix_array - 它们非常节省空间，并且有一些合理的可编程算法来生成它们，例如 Karkkainen 和 Sanders 的“简单线性工作后缀数组构造”

Answer 3

这是一个使用最简单后缀树的最长重复子串的简单实现。后缀树用这种方式很容易实现。

#include <iostream>
#include <vector>
#include <unordered_map>
#include <string>
using namespace std;

class Node
{
public:
    char ch;
    unordered_map<char, Node*> children;
    vector<int> indexes; //store the indexes of the substring from where it starts
    Node(char c):ch(c){}
};

int maxLen = 0;
string maxStr = "";

void insertInSuffixTree(Node* root, string str, int index, string originalSuffix, int level=0)
{
    root->indexes.push_back(index);

    // it is repeated and length is greater than maxLen
    // then store the substring
    if(root->indexes.size() > 1 && maxLen < level)
    {
        maxLen = level;
        maxStr = originalSuffix.substr(0, level);
    }

    if(str.empty()) return;

    Node* child;
    if(root->children.count(str[0]) == 0) {
        child = new Node(str[0]);
        root->children[str[0]] = child;
    } else {
        child = root->children[str[0]];
    }

    insertInSuffixTree(child, str.substr(1), index, originalSuffix, level+1);
}

int main()
{
    string str = "banana"; //"abcabcaacb"; //"banana";  //"mississippi";
    Node* root = new  Node('@');

    //insert all substring in suffix tree
    for(int i=0; i<str.size(); i++){
        string s = str.substr(i);
        insertInSuffixTree(root, s, i, s);
    }

    cout << maxLen << "->" << maxStr << endl;

    return 1;
}

/*
s = "mississippi", return "issi"
s = "banana", return "ana"
s = "abcabcaacb", return "abca"
s = "aababa", return "aba"
*/

Answer 4

LRS 问题最好使用后缀树或后缀数组来解决。两种方法都具有 O(n) 的最佳时间复杂度。

这是一个使用后缀数组的 LRS 问题的 O(nlog(n)) 解决方案。如果您有后缀数组的线性构造时间算法（这很难实现），我的解决方案可以改进为 O(n)。代码取自我的library。如果您想了解有关后缀数组如何工作的更多信息，请务必查看我的tutorials

/**
 * Finds the longest repeated substring(s) of a string.
 * 
 * Time complexity: O(nlogn), bounded by suffix array construction
 *
 * @author William Fiset, william.alexandre.fiset@gmail.com
 **/

import java.util.*;

public class LongestRepeatedSubstring {

  // Example usage
  public static void main(String[] args) {

    String str = "ABC$BCA$CAB";
    SuffixArray sa = new SuffixArray(str);
    System.out.printf("LRS(s) of %s is/are: %s\n", str, sa.lrs());

    str = "aaaaa";
    sa = new SuffixArray(str);
    System.out.printf("LRS(s) of %s is/are: %s\n", str, sa.lrs());

    str = "abcde";
    sa = new SuffixArray(str);
    System.out.printf("LRS(s) of %s is/are: %s\n", str, sa.lrs());

  }

}

class SuffixArray {

  // ALPHABET_SZ is the default alphabet size, this may need to be much larger
  int ALPHABET_SZ = 256, N;
  int[] T, lcp, sa, sa2, rank, tmp, c;

  public SuffixArray(String str) {    
    this(toIntArray(str));    
  }

  private static int[] toIntArray(String s) {   
    int[] text = new int[s.length()];   
    for(int i=0;i<s.length();i++)text[i] = s.charAt(i);   
    return text;    
  }

  // Designated constructor
  public SuffixArray(int[] text) {
    T = text;
    N = text.length;
    sa = new int[N];
    sa2 = new int[N];
    rank = new int[N];
    c = new int[Math.max(ALPHABET_SZ, N)];
    construct();
    kasai();
  }

  private void construct() {
    int i, p, r;
    for (i=0; i<N; ++i) c[rank[i] = T[i]]++;
    for (i=1; i<ALPHABET_SZ; ++i) c[i] += c[i-1];
    for (i=N-1; i>=0; --i) sa[--c[T[i]]] = i;
    for (p=1; p<N; p <<= 1) {
      for (r=0, i=N-p; i<N; ++i) sa2[r++] = i;
      for (i=0; i<N; ++i) if (sa[i] >= p) sa2[r++] = sa[i] - p;
      Arrays.fill(c, 0, ALPHABET_SZ, 0);
      for (i=0; i<N; ++i) c[rank[i]]++;
      for (i=1; i<ALPHABET_SZ; ++i) c[i] += c[i-1];
      for (i=N-1; i>=0; --i) sa[--c[rank[sa2[i]]]] = sa2[i];
      for (sa2[sa[0]] = r = 0, i=1; i<N; ++i) {
          if (!(rank[sa[i-1]] == rank[sa[i]] &&
              sa[i-1]+p < N && sa[i]+p < N &&
              rank[sa[i-1]+p] == rank[sa[i]+p])) r++;
          sa2[sa[i]] = r;
      } tmp = rank; rank = sa2; sa2 = tmp;
      if (r == N-1) break; ALPHABET_SZ = r + 1;
    }
  }

  // Use Kasai algorithm to build LCP array
  private void kasai() {
    lcp = new int[N];
    int [] inv = new int[N];
    for (int i = 0; i < N; i++) inv[sa[i]] = i;
    for (int i = 0, len = 0; i < N; i++) {
      if (inv[i] > 0) {
        int k = sa[inv[i]-1];
        while( (i + len < N) && (k + len < N) && T[i+len] == T[k+len] ) len++;
        lcp[inv[i]-1] = len;
        if (len > 0) len--;
      }
    }
  }

  // Finds the LRS(s) (Longest Repeated Substring) that occurs in a string.
  // Traditionally we are only interested in substrings that appear at
  // least twice, so this method returns an empty set if this is not the case.
  // @return an ordered set of longest repeated substrings
  public TreeSet <String> lrs() {

    int max_len = 0;
    TreeSet <String> lrss = new TreeSet<>();

    for (int i = 0; i < N; i++) {
      if (lcp[i] > 0 && lcp[i] >= max_len) {

        // We found a longer LRS
        if ( lcp[i] > max_len )
          lrss.clear();

        // Append substring to the list and update max
        max_len = lcp[i];
        lrss.add( new String(T, sa[i], max_len) );

      }
    }

    return lrss;

  }

  public void display() {
    System.out.printf("-----i-----SA-----LCP---Suffix\n");
    for(int i = 0; i < N; i++) {
      int suffixLen = N - sa[i];
      String suffix = new String(T, sa[i], suffixLen);
      System.out.printf("% 7d % 7d % 7d %s\n", i, sa[i],lcp[i], suffix );
    }
  }

}

Answer 5

public class LongestSubString {

    public static void main(String[] args) {
        String s = findMaxRepeatedString("ssssssssssss this is a ddddddd word with iiiiiiiiiis and loads of these are ppppppppppppps");
        System.out.println(s);
    }

    private static String findMaxRepeatedString(String s) {
        Processor p = new Processor();
        char[] c = s.toCharArray();
        for (char ch : c) {
            p.process(ch);
        } 
        System.out.println(p.bigger());
        return new String(new char[p.bigger().count]).replace('\0', p.bigger().letter);
    }

    static class  CharSet {
        int count;
        Character letter;
        boolean isLastPush;

        boolean assign(char c) {
            if (letter == null) {
                count++;
                letter = c;
                isLastPush = true;
                return true;
            }
            return false;
        }

        void reassign(char c) {
            count = 1;
            letter = c;
            isLastPush = true;
        }

        boolean push(char c) {
            if (isLastPush && letter == c) {
                count++;
                return true;
            }
            return false;
        }

        @Override
        public String toString() {
            return "CharSet [count=" + count + ", letter=" + letter + "]";
        }

    }

    static class  Processor {

        Character previousLetter = null;
        CharSet set1 = new CharSet();
        CharSet set2 = new CharSet();

        void process(char c) {
            if ((set1.assign(c)) || set1.push(c)) {
                set2.isLastPush = false;
            } else if ((set2.assign(c)) || set2.push(c)) {
                set1.isLastPush = false;                
            } else {
                set1.isLastPush = set2.isLastPush = false;
                smaller().reassign(c);
            }
        }       

        CharSet smaller() {
            return set1.count < set2.count ? set1 : set2;
        }

        CharSet bigger() {
            return set1.count < set2.count ? set2 : set1;
        }

    }   
}

Answer 6

我有一个面试，我需要解决这个问题。这是我的解决方案：

public class FindLargestSubstring {

public static void main(String[] args) {
    String test = "ATCGATCGA";
    System.out.println(hasRepeatedSubString(test));
}

private static String hasRepeatedSubString(String string) {
    Hashtable<String, Integer> hashtable = new Hashtable<>();
    int length = string.length();
    for (int subLength = length - 1; subLength > 1; subLength--) {
        for (int i = 0; i <= length - subLength; i++) {
            String sub = string.substring(i, subLength + i);
            if (hashtable.containsKey(sub)) {
                return sub;
            } else {
                hashtable.put(sub, subLength);
            }
        }
    }
    return "No repeated substring!";
}}

Answer 7

影响性能的因素太多了，我们无法仅用您提供的信息来回答这个问题。 （操作系统、语言、内存问题、代码本身）

如果您只是在寻找算法效率的数学分析，您可能想要改变问题。

编辑

当我提到“内存问题”和“代码”时，我没有提供所有细节。您将要分析的字符串的长度是一个重要因素。此外，代码不能单独运行——它必须位于程序内部才能发挥作用。该程序的哪些特征会影响该算法的使用和性能？

基本上，在您有实际情况要测试之前，您无法进行性能调整。您可以对什么可能表现最好做出非常有根据的猜测，但在您拥有真实数据和真实代码之前，您永远无法确定。