取List集合中最大的日期, 可以用Date max = Collections.max(dateList);,  传入一个日期集合, 就可以获取, 工作中有这个需求, 就查找到这个, 代码如下

		} else {

			/** 获取此专题下的所有内容的最新时间 */
			Long featureId = this.communityFeatureMapper.selectFeatureIdByLabelId(labelId);
			List<CommunityFeatureRelation> communityFeatureRelationList = this.communityFeatureRelationMapper.selectByFeatureId(featureId);
			List<Date> dateList = Lists.newArrayList();
			for (CommunityFeatureRelation communityFeatureRelation : communityFeatureRelationList) {
				CommunityProduct communityProduct =	this.communityProductMapper.selectContentIdByProductIdAndType(communityFeatureRelation.getProductId(),BaseConstans.ARTICLE_YIFABU);
				CommunityArticle communityArticle = this.communityArticleMapper.selectByPrimaryKey(communityProduct.getContentId());
				dateList.add(communityArticle.getReleaseTime());
			}
			if (!CollectionUtils.isEmpty(dateList)) {
				Date max = Collections.max(dateList);
				/** 内容->添加专题-此专题下的合伙人getReleaseTime 更新, 若此专题下不存在合伙人, 则不更新 */
				Long productId = this.communityProductMapper.selectIdByContentIdAndType(featureId, BaseConstans.FOUR);
				CommunityPartner communityPartner = this.communityPartnerMapper.selectByPartnerId(productId);
				if (!StringUtils.isEmpty(communityPartner)) {
					communityPartner.setCreateTime(max);
					communityPartnerMapper.updateByPrimaryKeySelective(communityPartner);
				}
			}


			 CommunityProduct communityProduct = this.communityProductMapper.selectContentIdByProductIdAndType(prodId, proType);