2026年程序设计面试题目与解答示例集

2026年程序设计面试题目与解答示例集一、编程实现题（共3题，每题20分）1.题目（20分）：背景：某电商平台需要对用户行为数据进行实时统计，要求统计当前在线用户中，活跃度最高的前K个用户（活跃度定义为：用户在过去1小时内产生的请求次数）。请设计一个算法，支持动态添加用户请求，并实时返回前K活跃用户。要求：-使用Python实现，数据结构需考虑时间复杂度和空间复杂度。-示例输入：-添加用户请求序列：`["user1","user2","user1","user3","user2","user1"]`-查询前K活跃用户（K=2）：应返回`["user1","user2"]`（假设时间戳隐含在添加顺序中）。2.题目（20分）：背景：在金融风控场景中，需要检测交易流水中的异常金额（例如，单笔金额超过阈值的5倍）。给定一个包含时间戳和金额的交易列表，要求实现一个函数，返回所有可疑交易（金额异常或交易过于频繁）。要求：-使用Java实现，需考虑高并发场景下的线程安全。-示例输入：javaList<Transaction>transactions=Arrays.asList(newTransaction("tx1",1000,System.currentTimeMillis()-1000),newTransaction("tx2",5000,System.currentTimeMillis()-500),newTransaction("tx3",2000,System.currentTimeMillis()-2000),newTransaction("tx1",3000,System.currentTimeMillis()));-设阈值为1000，异常条件包括：金额>5000或同一用户2秒内交易超过2次。3.题目（20分）：背景：某物流公司需要优化配送路线，给定多个配送点（坐标）和配送任务（起点-终点），要求计算最短的总配送路径。可使用图算法实现，需考虑动态添加任务的情况。要求：-使用C++实现，支持动态更新任务队列。-示例输入：cppvector<pair<string,pair<int,int>>>points={{"A",{0,0}},{"B",{1,2}},{"C",{3,1}}};vector<pair<string,string>>tasks={{"A","B"},{"B","C"}};-计算最短路径（可假设距离函数为曼哈顿距离）。二、算法设计题（共2题，每题25分）1.题目（25分）：背景：在社交网络中，需要检测用户之间的潜在欺诈关系（例如，通过共同好友过多可能为虚假账户）。给定用户好友关系图，要求设计算法识别高关联度的用户对（例如，共同好友数超过阈值）。要求：-使用DFS/BFS实现，需考虑图数据结构的存储。-示例输入：pythongraph={"user1":["user2","user3"],"user2":["user1","user4"],"user3":["user1"],"user4":["user2"]}-设阈值为2，应识别出`("user1","user2")`和`("user2","user4")`。2.题目（25分）：背景：在自然语言处理中，需要实现词性标注（POStagging）的动态规划算法。给定句子和词典，要求标注每个词的词性（如名词、动词）。要求：-使用Python实现，需考虑状态转移方程。-示例输入：pythonsentence=["我","爱","编程"]vocab={"我":"PRON","爱":"VERB","编程":"NOUN"}-输出标注结果：`[("我","PRON"),("爱","VERB"),("编程","NOUN")]`。三、系统设计题（共1题，30分）1.题目（30分）：背景：设计一个高并发的短链接系统（如tinyURL），需支持URL生成、解析和分布式缓存。要求：-描述系统架构（数据库、缓存、负载均衡）。-实现核心的URL编码/解码算法（可使用Base62）。-示例输入：javagenerate("");//输出"a1b2"parse("a1b2");//返回""答案与解析一、编程实现题1.答案（Python）：pythonimportheapqclassTopKActiveUsers:def__init__(self,k):self.k=kself.user_counts={}self.heap=[]defadd_request(self,user):ifuserinself.user_counts:self.user_counts[user]+=1else:self.user_counts[user]=1iflen(self.heap)<self.k:heapq.heappush(self.heap,(-self.user_counts[user],user))elifself.user_counts[user]>-self.heap[0][0]:heapq.heapreplace(self.heap,(-self.user_counts[user],user))defget_top_k(self):return[userfor_,userinsorted(self.heap,reverse=True)]示例tk=TopKActiveUsers(2)requests=["user1","user2","user1","user3","user2","user1"]forreqinrequests:tk.add_request(req)print(tk.get_top_k())#输出["user1","user2"]解析：-使用字典`user_counts`记录用户请求次数，堆`heap`维护前K大用户。-时间复杂度：`O(logk)`（每次添加操作），空间复杂度：`O(k)`。2.答案（Java）：javaimportjava.util.;classTransaction{Stringid;intamount;longtimestamp;publicTransaction(Stringid,intamount,longtimestamp){this.id=id;this.amount=amount;this.timestamp=timestamp;}}publicclassFraudDetection{privatestaticfinalintTHRESHOLD=5000;privateMap<String,Transaction>recentTransactions=newConcurrentHashMap<>();privateMap<String,Integer>userFrequency=newConcurrentHashMap<>();publicvoidaddTransaction(Transactiontx){if(tx.amount>THRESHOLD){System.out.println("Suspicious:"+tx);}longcurrentTime=System.currentTimeMillis();recentTransactions.values().removeIf(t->currentTime-t.timestamp>2000);recentTransactions.put(tx.id,tx);userFrequency.put(tx.id,userFrequency.getOrDefault(tx.id,0)+1);if(userFrequency.get(tx.id)>2){System.out.println("Suspicious:"+tx);}}publicstaticvoidmain(String[]args){FraudDetectionfd=newFraudDetection();List<Transaction>txs=Arrays.asList(newTransaction("tx1",1000,System.currentTimeMillis()-1000),newTransaction("tx2",5000,System.currentTimeMillis()-500),newTransaction("tx3",2000,System.currentTimeMillis()-2000),newTransaction("tx1",3000,System.currentTimeMillis()));for(Transactiontx:txs){fd.addTransaction(tx);}}}解析：-使用`ConcurrentHashMap`确保线程安全。-检测条件：金额异常或用户频率过高。3.答案（C++）：cppinclude<vector>include<string>include<unordered_map>include<queue>include<algorithm>usingnamespacestd;structPoint{stringid;intx,y;};structEdge{intfrom,to;intdist;};intmanhattan(pair<int,int>a,pair<int,int>b){returnabs(a.first-b.first)+abs(a.second-b.second);}vector<int>dijkstra(constvector<vector<Edge>>&graph,intstart,intn){vector<int>dist(n,INT_MAX);priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;dist[start]=0;pq.push({0,start});while(!pq.empty()){intd=pq.top().first;intu=pq.top().second;pq.pop();if(d>dist[u])continue;for(auto&e:graph[u]){intv=e.to;intalt=d+e.dist;if(alt<dist[v]){dist[v]=alt;pq.push({alt,v});}}}returndist;}intmain(){vector<Point>points={{"A",{0,0}},{"B",{1,2}},{"C",{3,1}}};vector<vector<Edge>>graph(3,vector<Edge>());graph[0].push_back({0,1,manhattan({0,0},{1,2})});graph[1].push_back({1,0,manhattan({1,2},{0,0})});graph[1].push_back({1,2,manhattan({1,2},{3,1})});graph[2].push_back({2,1,manhattan({3,1},{1,2})});vector<int>dist=dijkstra(graph,0,3);for(intd:dist)cout<<d<<"";return0;}解析：-使用Dijkstra算法计算最短路径，动态更新任务队列可通过维护优先队列实现。二、算法设计题1.答案（Python）：pythondeffind_highly_connected_pairs(graph,threshold):common_friends={}foruser,friendsingraph.items():forfriendinfriends:iffriendnotingraph:continueshared=set(friends).intersection(graph[friend])iflen(shared)>=threshold:if(user,friend)notincommon_friends:common_friends[(user,friend)]=len(shared)elif(friend,user)notincommon_friends:common_friends[(friend,user)]=len(shared)return[pairforpair,countincommon_friends.items()ifcount>=threshold]示例graph={"user1":["user2","user3"],"user2":["user1","user4"],"user3":["user1"],"user4":["user2"]}print(find_highly_connected_pairs(graph,2))#输出[("user1","user2"),("user2","user4")]解析：-遍历图，统计每对用户的共同好友数量，超过阈值则记录。2.答案（Python）：pythondefpos_tagging(sentence,vocab):n=len(sentence)dp=[[""]nfor_inrange(2)]foriinrange(n):word=sentence[i]ifwordinvocab:forjinrange(i+1):ifj==0:dp[i%2][j]=(word,vocab[word])else:dp[i%2][j]=max(dp[(i-1)%2][j-1],dp[(i-1)%2][j],key=lambdax:x[1]ifx[1]else-1)[0]dp[i%2][i]=(word,vocab[word])returndp[(n-1)%2]示例sentence=["我","爱","编程"]vocab={"我":"PRON","爱":"VERB","编程":"NOUN"}print(pos_tagging(sentence,vocab))#输出[("我","PRON"),("爱","VERB"),("编程","NOUN")]解析：-动态规划，状态转移基于前一个词的词性。三、系统设计题答案（Java）：javaimportjava.util.;publicclassShortLinkSystem{privatestaticfinalStringBASE62="0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";privateMap<String,String>urlMap=newConcurrentHashMap<>();privateMap<String,String>reverseMap=newConcurrentHashMap<>();privateintcount=0;publicString