Java字符串模式识别

Question

我有一个大约一千个字符长的字符串，由 L、T 和 A 组成。我很确定其中有一个简单的模式，我想知道是否有任何快速简便的方法可以找到它。该字符串会发生变化，因此这不仅仅是一次性的。

我正在寻找的模式是，例如，如果字符串是

LLLLLLLLAATAALLLLALLLLLLAATAALLLATLLLLLAATAALLAALLLLLAATAALL

子字符串LLLLLAATAALL在此字符串中重复4次。我想搜索这样的子字符串，但我不知道它们在哪里开始、结束、有多少个以及它们在主字符串中的长度。

如果 Java 中有任何工具可以查找此类内容，我们将不胜感激。

恩特

Answer 1

好的，所以我从 此处 获取代码并对其进行调整以跟踪并打印最长重复子串。只需使用 JUnit 运行它即可。

/* Copyright (c) 2010 the authors listed at the following URL, and/or
 the authors of referenced articles or incorporated external code:
 http://en.literateprograms.org/Suffix_tree_(Java)?action=history&offset=20100123220137

 Permission is hereby granted, free of charge, to any person obtaining
 a copy of this software and associated documentation files (the
 "Software"), to deal in the Software without restriction, including
 without limitation the rights to use, copy, modify, merge, publish,
 distribute, sublicense, and/or sell copies of the Software, and to
 permit persons to whom the Software is furnished to do so, subject to
 the following conditions:

 The above copyright notice and this permission notice shall be
 included in all copies or substantial portions of the Software.

 THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
 EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
 MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
 IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY
 CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT,
 TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE
 SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

 Retrieved from: http://en.literateprograms.org/Suffix_tree_(Java)?oldid=16641
 */

import java.util.ArrayList;
import java.util.Collection;
import java.util.List;
import org.junit.Test;

public class SuffixTree {
    @Test
    public void sampleUsage() {

        AbstractSuffixTree tree = new SimpleSuffixTree(
                "LLLLLLLLAATAALLLLALLLLLLAATAALLLATLLLLLAATAALLAALLLLLAATAALL");
        System.out.println("Longest repeating substring "
                + tree.best.printResult() + " repetitions=" + tree.best.visits
                + " length=" + tree.best.stringDepth);
    }
}

abstract class AbstractSuffixTree {

    SuffixTreeNode best;
    String text = null;
    SuffixTreeNode root = null;
    int inputAlphabetSize = -1;

    AbstractSuffixTree(String text) {
        if (text.length() > 0 && text.charAt(text.length() - 1) == '
输出：

最长重复子串

  LLLLLLAATAALLL 重复次数=2 长度=14

编辑：回复您的评论。
目前，我只是跟踪最长的重复 SuffixTreeNode（它是 AbstractSuffixTree 中的一个字段）。您可以修改它，以便它跟踪节点的 SortedQueue，按其 stringDepth 排序。
) {
            this.text = text;
        } else {
            this.text = text + "$";
        }
    }

}

class SimpleSuffixTree extends AbstractSuffixTree {

    public SimpleSuffixTree(String text) {
        super(text);
        constructTree();
    }

    private void constructTree() {
        super.root = new SuffixTreeNode(this);
        best = root;
        char[] s = super.text.toCharArray();
        for (int i = 0; i < s.length; i++) {
            List<String> suffixList = new ArrayList<String>();
            for (int k = i; k < s.length; k++) {
                suffixList.add(s[k] + "");
            }
            super.root.addSuffix(suffixList, i + 1);
        }
    }

}

class CompactSuffixTree extends AbstractSuffixTree {

    public CompactSuffixTree(SimpleSuffixTree simpleSuffixTree) {
        super(simpleSuffixTree.text);
        super.root = compactNodes(simpleSuffixTree.root, 0);
        super.best = simpleSuffixTree.best;
    }

    private SuffixTreeNode compactNodes(SuffixTreeNode node, int nodeDepth) {
        node.nodeDepth = nodeDepth;
        for (SuffixTreeNode child : node.children) {
            while (child.children.size() == 1) {
                SuffixTreeNode grandchild = child.children.iterator().next();
                child.incomingEdge.label += ", "
                        + grandchild.incomingEdge.label;
                child.stringDepth += grandchild.incomingEdge.label.length();
                child.children = grandchild.children;
                // for (SuffixTreeNode grandchild : child.children)
                grandchild.parent = node;
            }
            child = compactNodes(child, nodeDepth + 1);
        }
        return node;
    }

}

class SuffixTreeNode {

    AbstractSuffixTree tree;
    SuffixTreeEdge incomingEdge = null;
    int nodeDepth = -1;
    int label = -1;
    Collection<SuffixTreeNode> children = null;
    SuffixTreeNode parent = null;
    int stringDepth;
    int id = 0;
    public static int c;
    public int visits = 1;

    public SuffixTreeNode(AbstractSuffixTree tree, SuffixTreeNode parent,
            String incomingLabel, int depth, int label, int id) {
        children = new ArrayList<SuffixTreeNode>();
        incomingEdge = new SuffixTreeEdge(incomingLabel, label);
        nodeDepth = depth;
        this.label = label;
        this.parent = parent;
        stringDepth = parent.stringDepth + incomingLabel.length();
        this.id = id;
        this.tree = tree;
    }

    public SuffixTreeNode(AbstractSuffixTree tree) {
        children = new ArrayList<SuffixTreeNode>();
        nodeDepth = 0;
        label = 0;
        this.tree = tree;
    }

    public void addSuffix(List<String> suffix, int pathIndex) {
        SuffixTreeNode insertAt = this;
        insertAt = search(this, suffix);
        insert(insertAt, suffix, pathIndex);
    }

    private SuffixTreeNode search(SuffixTreeNode startNode, List<String> suffix) {
        if (suffix.isEmpty()) {
            throw new IllegalArgumentException(
                    "Empty suffix. Probably no valid simple suffix tree exists for the input.");
        }
        Collection<SuffixTreeNode> children = startNode.children;
        for (SuffixTreeNode child : children) {
            if (child.incomingEdge.label.equals(suffix.get(0))) {
                suffix.remove(0);
                child.visits++;
                if (child.visits > 1
                        && child.stringDepth > tree.best.stringDepth) {
                    tree.best = child;
                }
                if (suffix.isEmpty()) {
                    return child;
                }
                return search(child, suffix);
            }
        }
        return startNode;
    }

    private void insert(SuffixTreeNode insertAt, List<String> suffix,
            int pathIndex) {
        for (String x : suffix) {
            SuffixTreeNode child = new SuffixTreeNode(tree, insertAt, x,
                    insertAt.nodeDepth + 1, pathIndex, id);
            insertAt.children.add(child);
            insertAt = child;
        }
    }

    public String toString() {
        StringBuilder result = new StringBuilder();
        String incomingLabel = this.isRoot() ? "" : this.incomingEdge.label;
        for (int i = 1; i <= this.nodeDepth; i++)
            result.append("\t");
        if (this.isRoot()) {
            c = 1;
            this.id = 1;
        } else {
            this.id = c;
            result.append(this.parent.id + " -> ");
            result.append(this.id + "[label=\"" + incomingLabel + "\"]" + "("
                    + visits + "," + (stringDepth) + ")" + ";\n");
        }
        for (SuffixTreeNode child : children) {
            c++;
            child.id = c;
            result.append(child.toString());
        }
        return result.toString();
    }

    public String printResult() {
        if (parent == null) {
            return "";
        } else {
            return this.parent.printResult() + this.incomingEdge.label;
        }
    }

    public boolean isRoot() {
        return this.parent == null;
    }

    public boolean isLeaf() {
        return children.size() == 0;
    }

}

class SuffixTreeEdge {

    String label = null;
    @SuppressWarnings("unused")
    private int branchIndex = -1;

    public SuffixTreeEdge(String label, int branchIndex) {
        this.label = label;
        this.branchIndex = branchIndex;
    }

}

输出：

最长重复子串
LLLLLLAATAALLL 重复次数=2 长度=14

编辑：回复您的评论。

目前，我只是跟踪最长的重复 SuffixTreeNode（它是 AbstractSuffixTree 中的一个字段）。您可以修改它，以便它跟踪节点的 SortedQueue，按其 stringDepth 排序。

Answer 2

您将从以下代码中了解如何对字符串“hi”进行计数；

公共 int countHi(String str) {

int count = 0,i = str.length() - 1;
字符串目标=“嗨”；

for(int j = 0;j < i;j++)
{
if(str.substring(j,j+2).equals(目标))
计数++；}
返回计数；
}