技术选型
DFA实现原理
DFA全称为：Deterministic Finite Automaton,即确定有穷自动机。
存储：一次性的把所有的敏感词存储到了多个map中，就是下图表示这种结构
敏感词：冰毒、大麻、大坏蛋
工具类
最下面的main方法是测试用的，实际上线可以删掉，直接调用方法即可
import java.util.*;public class SensitiveWordUtil {public static Map<String, Object> dictionaryMap = new HashMap<>();/*** 生成关键词字典库* @param words* @return*/public static void initMap(Collection<String> words) {if (words == null) {System.out.println("敏感词列表不能为空");return ;}// map初始长度words.size()，整个字典库的入口字数(小于words.size()，因为不同的词可能会有相同的首字)Map<String, Object> map = new HashMap<>(words.size());// 遍历过程中当前层次的数据Map<String, Object> curMap = null;Iterator<String> iterator = words.iterator();while (iterator.hasNext()) {String word = iterator.next();curMap = map;int len = word.length();for (int i =0; i < len; i++) {// 遍历每个词的字String key = String.valueOf(word.charAt(i));// 当前字在当前层是否存在, 不存在则新建, 当前层数据指向下一个节点, 继续判断是否存在数据Map<String, Object> wordMap = (Map<String, Object>) curMap.get(key);if (wordMap == null) {// 每个节点存在两个数据: 下一个节点和isEnd(是否结束标志)wordMap = new HashMap<>(2);wordMap.put("isEnd", "0");curMap.put(key, wordMap);}curMap = wordMap;// 如果当前字是词的最后一个字，则将isEnd标志置1if (i == len -1) {curMap.put("isEnd", "1");}}}dictionaryMap = map;}/*** 搜索文本中某个文字是否匹配关键词* @param text* @param beginIndex* @return*/private static int checkWord(String text, int beginIndex) {if (dictionaryMap == null) {throw new RuntimeException("字典不能为空");}boolean isEnd = false;int wordLength = 0;Map<String, Object> curMap = dictionaryMap;int len = text.length();// 从文本的第beginIndex开始匹配for (int i = beginIndex; i < len; i++) {String key = String.valueOf(text.charAt(i));// 获取当前key的下一个节点curMap = (Map<String, Object>) curMap.get(key);if (curMap == null) {break;} else {wordLength ++;if ("1".equals(curMap.get("isEnd"))) {isEnd = true;}}}if (!isEnd) {wordLength = 0;}return wordLength;}/*** 获取匹配的关键词和命中次数* @param text* @return*/public static Map<String, Integer> matchWords(String text) {Map<String, Integer> wordMap = new HashMap<>();int len = text.length();for (int i = 0; i < len; i++) {int wordLength = checkWord(text, i);if (wordLength > 0) {String word = text.substring(i, i + wordLength);// 添加关键词匹配次数if (wordMap.containsKey(word)) {wordMap.put(word, wordMap.get(word) + 1);} else {wordMap.put(word, 1);}i += wordLength - 1;}}return wordMap;}public static void main(String[] args) {List<String> list = new ArrayList<>();list.add("法轮");list.add("法轮功");list.add("冰毒");initMap(list);String content="我是一个好人，并不会卖冰毒，也不操练法轮功,我真的不卖冰毒";Map<String, Integer> map = matchWords(content);System.out.println(map);}
}
