18.8 suffix tree

给定一个字符串s 和 一个包含较短字符串的数组t，设计一个方法，根据t中的每一个较短的字符换，对s进行搜索。

思路：基于s建立后缀树，然后在后缀树中进行查找，返回所有可能的索引值。

import java.util.ArrayList;

public class Question {
    public static void main(String[] args) {
        String testString = "bibs";
        String[] stringList = {"b", "bs", "hi", "bb"};
        SuffixTree tree = new SuffixTree(testString);
        for (String s : stringList) {
            ArrayList<Integer> list = tree.search(s);
            if (list != null) {
                System.out.println(s + ": " + list.toString());
            }
        }
    }
}

/**

b: [0, 2]

bs: [2]

*/

import java.util.ArrayList;

public class SuffixTree {
    SuffixTreeNode root = new SuffixTreeNode();
    
    public SuffixTree(String s) {
        for (int i = 0; i < s.length(); i++) {
            String suffix = s.substring(i);
            root.insertString(suffix, i);
        }
    }
    
    public ArrayList<Integer> search(String s) {
        return root.search(s);
    }
}

package Question18_8;

import java.util.ArrayList;
import java.util.HashMap;

public class SuffixTreeNode {
    HashMap<Character, SuffixTreeNode> children = new HashMap<Character, SuffixTreeNode>();
    
    char value;
    ArrayList<Integer> indexes = new ArrayList<Integer>();
    public SuffixTreeNode() { }
    
    public void insertString(String s, int index) {
        indexes.add(index);
        if (s != null && s.length() > 0) {
            value = s.charAt(0);
            SuffixTreeNode child = null;
            if (children.containsKey(value)) {
                child = children.get(value);
            } else {
                child = new SuffixTreeNode();
                children.put(value, child);
            }
            String remainder = s.substring(1);
            child.insertString(remainder, index);
        }
    }
    
    public ArrayList<Integer> search(String s) {
        if (s == null || s.length() == 0) {
            return indexes;
        } else {
            char first = s.charAt(0);
            if (children.containsKey(first)) {
                String remainder = s.substring(1);
                return children.get(first).search(remainder);
            }
        }
        return null;
    }

}