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There is a integer array which has the following features: The numbers in adjacent positions is different. A[0] We define a position P is a peek if:A[P] > A[P-1] A[P] > A[P+1]Find a peak element in the this array. Return the index of the peak.Given[1, 2, 1, 3, 4, 5, 7, 6]Return index 1 (which is number 2) or 6 (which is number 7)classSolution {/** * @parama:an integers array. * @return: Return any of peek positions. */ Public intFindpeak (int[] A) {//Write your code h

Divide-integers without using multiplication, division and mod operator.If It is overflow, return2147483647Given dividend = 100 and divisor = 9 , return 11 . Public classSolution {/** * @paramDividend The dividend *@paramdivisor the divisor *@returnThe result*/ Public intDivideintDividend,intdivisor) { //Write Your code here if(Divisor = = 0) return2147483647; if(Dividend = = 0) return0; if(Divisor = = 1) returndividend; if(Dividend

: Find how many characters in the beginning of a string str1 are characters that belong to the string str2function return: Returns the number of characters in the string str2 that are returned by the str1 at the beginning.Parameter description: STR1 is the lookup string, str2 is the string to find15. Function Name: strstrfunction prototypes: int strspn (const char *str1,const char *STR2)function function: Finds characters in string str1 that match exactly the characters in the STR2The function r

as an XML file, Next to the interface in the Catalog file, select the XML file, and then description fill in a nameThen, when you create a MAVEN project, you can select this directly, and when you choose, the contents of the native XML will be read directly, much faster than the Internet access.Finally give a little trick:Sometimes, in the unit's LAN Architecture Nexus Server, the server does not have access to the external network, so that can not be downloaded from the Internet through the Ag

/functionsIf [$#-ne 2];thenecho "At least input-argus:"Exit 1FiFor I in 12 13DoSCP-P22-RP $ [email protected] $i: ~ >/dev/null 2>1ssh-p22-t [email protected] $i SUDO/BIN/CP ~/$file $remotefile >/dev/nullIf [$?-eq 0];thenAction "SCP $file to $remotefile is OK"/bin/trueElseAction "SCP $file to $remotefile is failed"/bin/falseFidone , N Bsp , NB Sp , NB Sp , NB Sp , NB Sp , NB Sp

Permutations (the full permutation problem-the classic of backtracking)Given A collection of distinct numbers, return all possible permutations.For example,[1,2,3]The following permutations:[[1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]]The main idea: to find a sequence of the full array.Train of thought: When doing the permutation combination, the simulation of the process to get ideas. First select one of the elements in the collection as the first bit, then select an element in the remaining el

Given A string S , find the longest palindromic substring in S . The maximum length S of is assume, and there exists one unique longest palindromic substring.ExampleGiven the string = "abcdzdcab" , return "cdzdc" .ChallengeO (N2) time is acceptable. Can do it in O (n) time.N^2 procedure: Public classSolution {/** * @params Input String *@returnThe longest palindromic substring*/ Publicstring Longestpalindrome (string s) {//Write Your code here if(s = =NULL|| S.length () = = 0)

is considered to be index, delete } } returncount==N; }}2.DFSThe pre must be recorded in DFS to avoid a->b->a such traversal, which is considered a loop.classSolution { Public BooleanValidtree (intNint[] edges) { if(n==0)return false; List[] List=NewList[n]; for(inti=0;i) {List[i]=NewArraylist(); } for(int[] pair:edges) {list[pair[0]].add (pair[1]); list[pair[1]].add (pair[0]); } Boolean[] visited=New Boolean[n]; if(Dfs (list, visited, 0,-1))retur

and N;(1) Termination criteria: X and Y are invalid, the currently moving character is not the next character of the string;(2) Returns True if the end of the matching string has been reached(3) Temporarily replace the current movement to the coordinate character ' ", because to make a mark, so as not to repeat the reentry movement;(4) Call 4 recursive functions, respectively moving to the top and bottom around the current motion character, and determine;(5) If 4 recursive functions are false,

---restore content starts---Given a list of integers, which denote a permutation.Find the next permutation in ascending order.NoticeThe list may contains duplicate integers.Example[1,3,2,3]for, the next permutation is[1,3,3,2][4,3,2,1]for, the next permutation is[1,2,3,4] Public classSolution {/** * @paramnums:an array of integers *@return: Return Nothing (void), does not return anything, modify Nums in-place instead*/ Public int[] Nextpermutation (int[] nums) { //Write your code

Given an array of integers and a number k, the majority number is the number of the occurs of the size of the more than 1/k array.Find it.NoticeThere is a majority number in the array.ExampleGiven [3,1,2,3,2,3,3,4,4,4] k=3 and, return 3 .ChallengeO (n) time and O (k) Extra space Public classSolution {/** * @paramnums:a List of integers *@paramK:as described *@return: The majority number*/ Public intMajoritynumber (arraylistintk) {//Write your code if(Nums = =NULL|| Nums.si

Given an array of integers, find the numbers such that they add up to a specific target number.The function twoSum should return indices of the numbers such that they add up to the target, where index1 mus T is less than INDEX2. Please note that your returned answers (both Index1 and Index2) is not zero-based.NoticeYou may assume this each input would has exactly one solutionExamplenumbers= [2, 7, 11, 15] , target=9Return[1, 2]ChallengeEither of the following solutions is acceptable: O

Write a program to find the nth super ugly number.Super ugly numbers is positive numbers whose all prime factors is in the given prime list primes of size k . For example, is the sequence of the first and [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] super ugly numbers given primes = of [2, 7, 13, 19] size 4 .Notice 1is a super ugly number for any given primes. The given numbers in primes is in ascending order. 0 ExampleGiven n = 6 , primes = [2, 7, 13, 19] return13 Public c

Given 3*n + 1 numbers, every numbers occurs triple times except one, find it.ExampleGiven [1,1,2,3,3,3,2,2,4,1] return4ChallengeOne-pass, constant extra space. Public classSolution {/** * @paramA:an Integer Array *@return: An integer*/ Public intSinglenumberii (int[] A) {//Write your code here if(A = =NULL|| A.length = = 0) return0; int[] Count =New int[32]; for(inti = 0; I ){ for(intj = 0; J ) {Count[i]+ = (A[j] >> i) 1; Count[i]%= 3; } }

Given a set of distinct integers, return all possible subsets.Notice Elements in a subset must is in non-descending order. The solution set must not contain duplicate subsets. ExampleIf S = [1,2,3] , a solution is:[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], []]ChallengeCan do it in both recursively and iteratively?classSolution {/** * @params:a set of numbers. * @return: A List of lists. All valid subsets. */ Publicarraylistint[] nums) { //Write yo

data to go heavy, feel that this is problematic, such as:-1-1 1 1, 1-1-1 3 This is also an answer, but the above nine chapters to the program is really by the deduplication operation ....Looking closely at the nine chapters of the procedure, found:1,i is the first number, J is the second number, start is the third number, end is the fourth number2. The first two numbers are de-weighed and the calculated start and end are also de-weighed, and start and end are well understood, but for the first

*/ Public intFirstmissingpositive (int[] A) {//Write your code here if(A.length = = 0) return1; intn =a.length; for(intI =0;i){ while(A[i]! = i + 1){ if(A[i] N) Break; intTMP =A[i]; A[i]= A[tmp-1]; A[tmp-1] =tmp; } } for(intI =0;i) if(A[i]! = i+ 1) returnI+1; returnn + 1; }}Java CodeclassSolution:#@param A, a list of integers #@return An integer deffirstmissingpositive (Self, A):#Write your code here

result*/ Public intDivideintDividend,intdivisor) { //Write Your code here if(Divisor ==0) returnInteger.max_value; if(Divisor = =-1 Dividend = =integer.min_value)returnInteger.max_value; intCount = 0; LongPdividend = Math.Abs ((Long) dividend); LongPdivisor = Math.Abs ((Long) divisor); while(Pdividend >=pdivisor) { intCount1 = 1; Longsum =Pdivisor; while(sum + sum) pdividend) {Count1+=Count1; Sum+=sum; } Count+=Count1; Pdividend-=sum; } if(Divid

half of the minimum value of the second half of the maximum value, the subtraction is the answer, this seemingly difficult solution.At the time of the variable array, the minimum value in front of it is compared, and the current node can be regarded as the maximum value, the difference between the assumed maximum value and the minimum value of the previous one, and the profit obtained from the current node's sell, is the local maximum value.The maximum value of the local maximum is the overall

-target)?sum:tmpsum; if(Tmpdist = = 0){ returnTarget; } if(Tmpdist ) { left++; }Else{ Right--; } } } returnsum; }}Java CodeTotal time: 1504 MsclassSolution:"""@param numbers:give An array numbers of n integer @param target:an integer @return: Return the sum of T He three integers, the sum closest target. """ defthreesumclosest (self, Numbers, target):#Write your code here ifNumbers = =None:return0 Numlen=len (Numbers) number