背景
近期有几个业务方提出一需求，期望判断一个用户在短期内是否存在刷屏现象，出现后能对其做出限制，并上报。
刷屏定义：取出用户近期20条评论，如果有50%的评论是"相似"的，则认为该用户是在刷屏
相似定义：两条评论的字符串最小编辑距离 / 长串的长度 < 0.2,即两串的80%是相同的，则认为两串相似。
@Slf4j
public class SimpleBrushDetectionFilter implements ReviewFilter {
// Todo 参数可实时调
private int USER_RECENT_REVIEW_LIST_SIZE = 20;
private int SIMILARITY_THRESHOLD = 80;
private double BRUSH_THRESHOLD = 0.5;// 该值不允许低于0.5，否则会出现用户循环被ban
private int BAN_SECOND = 3600 * 24;//一天
private int LIST_EXPIRE_SECOND = 3600 * 24 * 3;//三天
@Override
public ReviewFilterModel filter(ReviewFilterModel reviewFilterModel) {
if (reviewFilterModel.isEnd()) {
return reviewFilterModel;
}
long userId = reviewFilterModel.getReviewInfo().getUserId();
if (userId <= 0) {
log.info("错误的userId {}", userId);
return reviewFilterModel;
}
BrowserRedisService banRedisInstance = BrowserRedisService
.getRedisService(RedisPrefix.REVIEW_SIMPLE_BRUSH_DETECTION_BAN);
String str = banRedisInstance.get("" + userId);
if (StrUtil.isNotBlank(str)
// BAN_SECOND的expire set非原子性。出错时需要额外判断一下
&& (System.currentTimeMillis() - Long.parseLong(str)) < BAN_SECOND * 1000) {
banReview(reviewFilterModel, userId);
return reviewFilterModel;
}
if (StrUtil.isNotBlank(str) && (System.currentTimeMillis() - Long.parseLong(str)) > BAN_SECOND * 1000) {
banRedisInstance.del("" + userId);
}
return simpleBrushDetect(reviewFilterModel);
}
private void banReview(ReviewFilterModel reviewFilterModel, long userId) {
log.info("user {} 疑似刷屏，限制发表评论", userId);
reviewFilterModel.setEnd(true);
reviewFilterModel.setPass(false);
reviewFilterModel.setReason("该用户疑似近期出现恶意刷屏，限制发表评论");
}
private ReviewFilterModel simpleBrushDetect(ReviewFilterModel reviewFilterModel) {
BrowserRedisService listRedisInstance = BrowserRedisService
.getRedisService(RedisPrefix.REVIEW_SIMPLE_BRUSH_DETECTION_LIST);
long userId = reviewFilterModel.getReviewInfo().getUserId();
List userRecentReview = listRedisInstance
.lrange("" + userId, 0, USER_RECENT_REVIEW_LIST_SIZE);
if (null == userRecentReview) {
// 将当前评论塞入队列中
listRedisInstance.rpush("" + userId, reviewFilterModel.getReviewInfo().getDocuments());
return reviewFilterModel;
}
userRecentReview.add(reviewFilterModel.getReviewInfo().getDocuments());
// 正确的暴力做法是，将20个串依次互相两两对比，但是这样复杂度太高了
// 这里采用一个取巧的方法，将20个串按字典序排序，然后依次左右对比，效果应该也可以接受
Collections.sort(userRecentReview);
int cnt = 0;
for (int i = 0; i < userRecentReview.size() - 1; i++) {
int similarity = towStringSimilarity(userRecentReview.get(i),
userRecentReview.get(i + 1));
if (similarity > SIMILARITY_THRESHOLD) {
cnt++;
}
}
if (cnt > BRUSH_THRESHOLD * USER_RECENT_REVIEW_LIST_SIZE) {
log.info("user {} 疑似刷屏，禁止发言{}秒", userId, BAN_SECOND);
BrowserRedisService banRedisInstance = BrowserRedisService
.getRedisService(RedisPrefix.REVIEW_SIMPLE_BRUSH_DETECTION_BAN);
banRedisInstance.set("" + userId, "" + System.currentTimeMillis());
banRedisInstance.expire("" + userId, BAN_SECOND);
// 为了避免用户禁言到期后再次触发逻辑，list中删除2/3的评论
listRedisInstance.ltrim("" + userId, -1, -USER_RECENT_REVIEW_LIST_SIZE / 3);
banReview(reviewFilterModel, userId);
}
// 将当前评论塞入队列中
listRedisInstance.rpush("" + userId, reviewFilterModel.getReviewInfo().getDocuments());
listRedisInstance.ltrim("" + userId, -1, -USER_RECENT_REVIEW_LIST_SIZE);
// 刷新整条list的过期时间
listRedisInstance.expire("" + userId, LIST_EXPIRE_SECOND);
return reviewFilterModel;
}
/**
* 返回两个字符串的相似度。 当某个串长度小于5的时候，认为其不构成可比性
*
* @return int [0,100]
*/
private static int towStringSimilarity(String word1, String word2) {
if (word1.length() < 5 || word2.length() < 5) {
return 0;
}
int distance = towStringMinDistance(word1, word2);
return 100
- distance / (word1.length() > word2.length() ? word1.length() : word2.length()) * 100;
}
/**
* 返回两条字符串的最短编辑距离,
*
* 即将word2转变成word1的最小操作次数。
*
* 采用二维动态规划实现，时间复杂度O(N^2)
*/
private static int towStringMinDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
if (m == 0) {
return n;
}
if (n == 0) {
return m;
}
int[][] f = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
f[i][0] = i;
}
for (int j = 0; j <= n; j++) {
f[0][j] = j;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
return f[m][n];
}
private static int min(int a, int b, int c) {
return (a > b ? (b > c ? c : b) : (a > c ? c : a));
}
}
