2025年計算機編程技術(shù)面試題庫1.代碼填空題（共5題，每題10分）題目1請?zhí)羁胀瓿梢韵翵avaScript代碼，使其能夠正確計算數(shù)組中所有正數(shù)的平均值。javascriptfunctioncalculatePositiveAverage(arr){letsum=0;letcount=0;for(leti=0;i<arr.length;i++){if(arr[i]>0){sum+=______;count++;}}returncount>0?______:0;}題目2請?zhí)羁胀瓿梢韵翽ython代碼，使其能夠反轉(zhuǎn)一個字符串。pythondefreverse_string(s):returns[______:______]題目3請?zhí)羁胀瓿梢韵翵ava代碼，使其能夠判斷一個整數(shù)是否為偶數(shù)。javapublicbooleanisEven(intnum){returnnum%2==______;}題目4請?zhí)羁胀瓿梢韵翪++代碼，使其能夠找出一個數(shù)組中的最大值。cppintfindMax(intarr[],intsize){intmax=arr[0];for(inti=1;i<size;i++){if(arr[i]>______){max=arr[i];}}returnmax;}題目5請?zhí)羁胀瓿梢韵翽ython代碼，使其能夠計算一個列表中所有元素的平方和。pythondefsquare_sum(lst):returnsum([x2forxin______])答案答案1javascriptfunctioncalculatePositiveAverage(arr){letsum=0;letcount=0;for(leti=0;i<arr.length;i++){if(arr[i]>0){sum+=arr[i];count++;}}returncount>0?sum/count:0;}答案2pythondefreverse_string(s):returns[::-1]答案3javapublicbooleanisEven(intnum){returnnum%2==0;}答案4cppintfindMax(intarr[],intsize){intmax=arr[0];for(inti=1;i<size;i++){if(arr[i]>max){max=arr[i];}}returnmax;}答案5pythondefsquare_sum(lst):returnsum([x2forxinlst])2.編程實現(xiàn)題（共5題，每題15分）題目1請用Python實現(xiàn)一個函數(shù)，該函數(shù)接收一個字符串，返回該字符串中所有重復字符的列表（不區(qū)分大小寫），每個字符只出現(xiàn)一次。題目2請用Java實現(xiàn)一個方法，該方法接收一個整數(shù)數(shù)組，返回一個新數(shù)組，新數(shù)組中的元素為原數(shù)組中每個元素的平方，且順序不變。題目3請用JavaScript實現(xiàn)一個函數(shù)，該函數(shù)接收一個正整數(shù)n，返回一個包含前n個斐波那契數(shù)的數(shù)組。題目4請用C++實現(xiàn)一個函數(shù)，該函數(shù)接收一個字符串，返回該字符串的長度，但不計算空格字符。題目5請用Python實現(xiàn)一個類，該類包含一個方法，能夠判斷一個給定的正整數(shù)是否為素數(shù)。答案答案1pythondeffind_duplicates(s):s_lower=s.lower()duplicates=[]forcharins_lower:ifs_lower.count(char)>1andcharnotinduplicates:duplicates.append(char)returnduplicates答案2javapublicint[]squareArray(int[]arr){int[]squared=newint[arr.length];for(inti=0;i<arr.length;i++){squared[i]=arr[i]*arr[i];}returnsquared;}答案3javascriptfunctionfibonacci(n){letfib=[0,1];for(leti=2;i<n;i++){fib[i]=fib[i-1]+fib[i-2];}returnfib.slice(0,n);}答案4cppintcustomLength(conststd::string&str){intlength=0;for(charc:str){if(c!=''){length++;}}returnlength;}答案5pythonclassPrimeChecker:defis_prime(self,num):ifnum<=1:returnFalseforiinrange(2,int(num0.5)+1):ifnum%i==0:returnFalsereturnTrue3.算法題（共5題，每題20分）題目1給定一個整數(shù)數(shù)組，返回數(shù)組中第三大的數(shù)。如果數(shù)組中沒有第三大的數(shù)，則返回最大的數(shù)。題目2給定一個字符串，請找到其中不重復的最長子串的長度。題目3給定一個二維網(wǎng)格，每個格子中的數(shù)字表示該格子的高度。找出一條從左上角到右下角的路徑，使得路徑上的數(shù)字之和最小。每次只能向下或向右移動。題目4給定一個非空鏈表，刪除鏈表的倒數(shù)第n個節(jié)點，并返回鏈表的頭節(jié)點。題目5給定一個包含n個整數(shù)的數(shù)組，判斷該數(shù)組是否可以劃分為至少兩個連續(xù)的子數(shù)組，每個子數(shù)組的和都相等。答案答案1pythondefthird_largest(nums):first,second,third=float('-inf'),float('-inf'),float('-inf')fornuminnums:ifnum>first:third,second,first=second,first,numelifnum>second:third,second=second,numelifnum>third:third=numreturnfirstifthird!=float('-inf')elsesecond答案2pythondeflength_of_longest_substring(s):char_map={}left=0max_length=0forrightinrange(len(s)):ifs[right]inchar_mapandchar_map[s[right]]>=left:left=char_map[s[right]]+1char_map[s[right]]=rightmax_length=max(max_length,right-left+1)returnmax_length答案3pythondefmin_path_sum(grid):ifnotgridornotgrid[0]:return0m,n=len(grid),len(grid[0])dp=[[0]*nfor_inrange(m)]dp[0][0]=grid[0][0]foriinrange(1,m):dp[i][0]=dp[i-1][0]+grid[i][0]forjinrange(1,n):dp[0][j]=dp[0][j-1]+grid[0][j]foriinrange(1,m):forjinrange(1,n):dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i][j]returndp[-1][-1]答案4pythondefremoveNthFromEnd(head,n):dummy=ListNode(0)dummy.next=headfirst=dummysecond=dummyfor_inrange(n+1):first=first.nextwhilefirst:first=first.nextsecond=second.nextsecond.next=second.next.nextreturndummy.next答案5pythondefcan_partition(nums):total_sum=sum(nums)iftotal_sum%2!=0:returnFalsetarget=total_sum//2dp=[False]*(target+1)dp[0]=Truefornuminnums:forjinrange(target,num-1,-1):dp[j]=dp[j]ordp[j-num]returndp[target]4.數(shù)據(jù)結(jié)構(gòu)題（共5題，每題20分）題目1請用Python實現(xiàn)一個棧類，該類支持push、pop和isEmpty操作。題目2請用Java實現(xiàn)一個隊列類，該類使用兩個棧實現(xiàn)隊列的操作。題目3請用C++實現(xiàn)一個二叉搜索樹類，該類支持插入和查找操作。題目4請用JavaScript實現(xiàn)一個雙向鏈表類，該類支持append、prepend和remove操作。題目5請用Python實現(xiàn)一個哈希表類，該類使用鏈表解決哈希沖突。答案答案1pythonclassStack:def__init__(self):self.items=[]defpush(self,item):self.items.append(item)defpop(self):ifnotself.isEmpty():returnself.items.pop()returnNonedefisEmpty(self):returnlen(self.items)==0答案2javaimportjava.util.Stack;publicclassQueueWithStacks{privateStack<Integer>stack1;privateStack<Integer>stack2;publicQueueWithStacks(){stack1=newStack<>();stack2=newStack<>();}publicvoidenqueue(intitem){stack1.push(item);}publicintdequeue(){if(stack2.isEmpty()){while(!stack1.isEmpty()){stack2.push(stack1.pop());}}returnstack2.pop();}publicbooleanisEmpty(){returnstack1.isEmpty()&&stack2.isEmpty();}}答案3cpp#include<iostream>usingnamespacestd;structTreeNode{intval;TreeNode*left;TreeNode*right;TreeNode(intx):val(x),left(nullptr),right(nullptr){}};classBST{public:TreeNode*insert(TreeNode*root,intval){if(!root)returnnewTreeNode(val);if(val<root->val)root->left=insert(root->left,val);elseroot->right=insert(root->right,val);returnroot;}boolsearch(TreeNode*root,intval){if(!root)returnfalse;if(root->val==val)returntrue;returnval<root->val?search(root->left,val):search(root->right,val);}};答案4javascriptclassListNode{constructor(val=0,prev=null,next=null){this.val=val;this.prev=prev;this.next=next;}}classDoublyLinkedList{constructor(){this.head=null;this.tail=null;}append(val){constnewNode=newListNode(val);if(!this.head){this.head=this.tail=newNode;}else{newNode.prev=this.tail;this.tail.next=newNode;this.tail=newNode;}}prepend(val){constnewNode=newListNode(val);if(!this.head){this.head=this.tail=newNode;}else{newNode.next=this.head;this.head.prev=newNode;this.head=newNode;}}remove(node){if(!node)return;if(node.prev){node.prev.next=node.next;}else{this.head=node.next;}if(node.next){node.next.prev=node.prev;}else{this.tail=node.prev;}}}答案5pythonclassListNode:def__init__(self,key=None,value=None):self.key=keyself.value=valueself.next=NoneclassHashTable:def__init__(self,size=100):self.size=sizeself.table=[None]*self.sizedef_hash(self,key):returnhash(key)%self.sizedefinsert(self,key,value):index=self._hash(key)node=self.table[index]ifnotnode:self.table[index]=ListNode(key,value)else:prev=Nonewhilenode:ifnode.key==key:node.value=valuereturnprev=nodenode=node.nextprev.next=ListNode(key,value)defget(self,key):index=self._hash(key)node=self.table[index]whilenode:ifnode.key==key:returnnode.valuenode=node.nextreturnNone5.面試綜合題（共5題，每題25分）題目1請設(shè)計一個算法，找出數(shù)組中重復次數(shù)最多的元素，并返回其重復次數(shù)。題目2請設(shè)計一個算法，將一個字符串中的所有字母移到字符串的左邊，所有非字母字符移到字符串的右邊，并保持字母和非字母字符的相對順序不變。題目3請設(shè)計一個算法，判斷一個給定的鏈表是否為回文鏈表。題目4請設(shè)計一個算法，找出二叉樹中所有路徑和等于給定數(shù)字的路徑。題目5請設(shè)計一個算法，實現(xiàn)一個簡單的LRU（最近最少使用）緩存，支持get和put操作。答案答案1pythondefmost_frequent(nums):count={}max_count=0most_freq=Nonefornuminnums:count[num]=count.get(num,0)+1ifcount[num]>max_count:max_count=count[num]most_freq=numreturnmax_count答案2pythondefreorderString(s):letters=[]non_letters=[]forcins:ifc.isalpha():letters.append(c)else:non_letters.append(c)return''.join(letters+non_letters)答案3pythondefisPalindrome(head):ifnotheadornothead.next:returnTrueslow=headfast=headprev=Nonewhilefastandfast.next:fast=fast.next.nextprev,slow.next=slow,prevslow=slow.nextiffast:slow=slow.nextwhileslow:ifslow.val!=prev.val:returnFalseslow=slow.nextprev=prev.nextreturnTrue答案4pythondefpathSum(root,sum):result=[]defdfs(node,current_sum,path):ifnotnode:returncurrent_sum+=node.valpath.append(node.val)ifcurrent_sum==sum:res