hackerrank vs leetcode

; int rob (vector int >num) { int best0=0; // indicates that the current houses NBSP;NBSP; int best1=0; // indicates that the current houses NBSP;NBSP; for int i=0;i int temp=best0; best0= max (best0,best1); // does not select the current houses, it is equal to the last selected or not selected maximum value NBSP;NBSP; best1= temp+num[i];// selected the current houses, the value can only be equal to the last not selected + The current houses of

Given an array, move the elements in the array to the right by the K position, where K is a non-negative number.Example 1:Input: [1,2,3,4,5,6,7] k = 3 output: [5,6,7,1,2,3,4] Explanation:[7,1,2,3,4,5,6][6,7,1,2,3,4,5][5,6,7,1,2,3,4]Example 2:Input: [-1,-100,3,99] k = 2 output: [3,99,-1,-100] Explanation: Rotate Right 1 steps: [99,-1,-100,3] Rotate Right 2 steps: [3,99,-1,-100]Description Think of as many solutions as possible, at least three different ways to solve the problem. Requ

You is a professional robber planning to rob houses along a street. Each house have a certain amount of money stashed, the only constraint stopping all from robbing each of the them are that Adjac ENT houses have security system connected and it would automatically contact the police if the adjacent houses were broken Into on the same night.Given a list of non-negative integers representing the amount of money in each house, determine the maximum amount of mone Y you can rob tonight without aler

the range on the left, it looks on the left;Otherwise, find on the rightIf the median is smaller than the left side, the right side is incremented, and the breakpoint is on the left:If Target is in the increment interval on the right, it is located on the right;Otherwise, find on the left.From the results of running, for small data, the solution is a little bit faster.‘‘‘Class Solution:# @param A, a list of integers# @param target, an integer to be searched# @return An integerdef search (self,

stack is not empty,Out of the stack, calculate if you need to update Max_so_farAlso, update start to the number of the stackIf the stack is empty,Start points to the next position of I‘‘‘Class Solution:# @param s, a string# @return An integerdef longestvalidparentheses (self, s):Start = 0Max_so_far = 0stack = []For i in Xrange (Len (s)):If s[i] = = "(":Stack.append (min (i, start))start = i + 1ElseIf Len (stack) >=1:Start = last = Stack.pop ()Max_so_far = Max (i-last+1, Max_so_far)Elsestart = i

for(inti=0;i) { at if(sch[i]==strch[index+i]) { - returns; - } - } -Str=str.substring (index+1); - in } - return NULL; to } + - the * //Method Two $ Public Staticstring StrStr2 (String haystack, string needle) {Panax Notoginseng if(needle.length () = = 0) - returnhaystack; the + for(inti = 0; I ) { A

with a new string that adds a substring, because this is a looping statement, and we will still use the original path for the next stop (no substring added).For (int Stop:mark[index]) { string substr = S.substr (index, stop-index); String NewPath = Path + (index = = 0? substr: "" + substr);AC Code:Class Solution {public:vectorThis problem is very comprehensive, it is difficult to combine DP and backtracking, especially the method of setting up DP array can be used f

) { - Chart2 =S.charat (start); in if(Need.containskey (T2)) { - if(Found.get (T2) >need.get (T2)) toFound.put (T2, Found.get (T2)-1); + Else - Break; the } * } $ if(start>end)Panax Notoginsengstart--; - if(end-startMinstart) { theMinstart =start; +Minend =end; A

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '15:3sumhttps://oj.leetcode.com/problems/3sum/Given an array S of n integers, is there elements a, B, C in S such that A + B + c = 0?Find all unique triplets in the array which gives the sum of zero.Note:Elements in a triplet (A,B,C) must is in non-descending order. (ie, a≤b≤c)The solution set must not contain duplicate triplets.For example, given array S = {-1 0 1 2-1-4},A Solution set is:(-1, 0, 1)(-1,-1, 2)===comments by dabay===First sort

inti = m-1, j = n-1, k = m+n-1;3 while(I>-1 j>-1)4a[k--] = A[i] > b[j]? a[i--]: b[j--];5 while(j>-1) 6a[k--] = b[j--];7}Answer Ideas :Since the trip will be covered, but the back is all 0, then from the back forward will not be covered.(even if the extreme situation, a is all less than B, the back of the 0 is just enough B, there will be no coverage)1, when I and J are not exhausted, from the back to the front, who is large who entered, and the corresponding pointer consum

Topic:Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.For example,Given N = 3 ,You should return the following matrix:[[1, 2, 3], [8, 9, 4], [7, 6, 5]]idea: The problem is the same as the spiral matrix model and solution. We are asked to write the 1~n^2 number into the matrix. We still maintain four variables, Rowbegin, Rowend, Colbegin, Colend. And then to define the boundary of the helix. What we need to note is that if n is 0, we should return an

5 6 7 8] Part if do the cut this. Therefore, for the longer array to is used as the basis for the first cut, a range check must is performed. It would is just easier to does it on the shorter array, which requires no checks whatsoever. Also, moving only on the shorter array gives a run-time complexity of O (log (min (N1, N2))) (edited as suggested by @baselRu s) B. The only edge case was when a cut falls on the 0th (first) or the 2Nth (last) position. For instance, if C2 = 2N2, then R2 = A2[2*N

) return "/";4. Note how to get the elements between the delimiters "/" and do the operation, we traverse through the path.while (I 5. First get the element of the queue header, then Pop_front (), convert to normal path.while (!stk.empty ()) { ret + = "/" + Stk.front (); Stk.pop_front (); }complexity: O (n) space is also O (n), stack sizeAC Code:Class Solution {public:string Simplifypath (string path) {if (path.size () = = 0) return ""; ListT

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '17:letter combinations of a Phone numberhttps://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/Given a digit string, return all possible letter combinations, the number could represent.A mapping of Digit to letters (just as on the telephone buttons) is given below.Input:digit string "23"Output: ["Ad", "AE", "AF", "BD", "Be", "BF", "CD", "CE", "CF"].===comments by dabay===Compare a typical recursive call.Each time a word is di

[mid] ]: +Start = Mid+1 A #end = Len (num)-1 at Else: - returnMidRecursive code: OJ Test passed runtime:51 ms1 classSolution:2 #@param num, a list of integers3 #@return An integer4 defFind (self,num,start,end):5 ifStart = =End:6 returnStart7 ifEnd = = Start + 1:8 ifNum[start] Num[end]:9 returnEndTen Else: One returnStart AMid = (start+end)/2 - ifNum[

Title Description:Given the arrays, write a function to compute their intersection.Example:Given nums1 = [1, 2, 2, 1] , nums2 = [2, 2] , return [2, 2] .Note: Each element of the result should appear as many times as it shows in both arrays. The result can be on any order. Follow up: What if the given array is already sorted? How would optimize your algorithm? What if nums1' s size is small compared to nums2' s size? Which algorithm is better? What if el

Topic:Related to question Excel Sheet Column TitleGiven a column title as appear in an Excel sheet, and return its corresponding column number.For example: 1, B, 2 C-3 ... 27 AA, Z idea: This problem and Excel Sheet Column title and is corresponding. This question asks us to convert the letters in Excel into numbers. Observe the law, we just need to figure out the corresponding number of each letter, iterative increase can be, note that the Excel table alphanumeric corresponden

]! = '. ' and Board[ii*3+row][jj*3+col] in s:S.remove (Board[ii*3+row][jj*3+col])return sdef solveSudoku2 (board, position):I, j = positionif i = = 9:Return TrueIf board[i][j] = = '. ':Nums = valid_nums (board, position)For N in Nums:BOARD[I][J] = nIf SOLVESUDOKU2 (board, next_position (position)) is True:Return TrueBOARD[I][J] = '. 'ElseReturn SOLVESUDOKU2 (board, next_position (position))SOLVESUDOKU2 (board, (0, 0))def print_board (board):Print "-" * 30For row in board:For x in row:Print "%s"%

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '35:search Insert Positionhttps://oj.leetcode.com/problems/search-insert-position/Given a sorted array and a target value, return the index if the target is found.If not, return the index where it would is if it were inserted in order.Assume no duplicates in the array.Here is few examples.[1,3,5,6], 5→2[1,3,5,6], 2→1[1,3,5,6], 7→4[1,3,5,6], 0→0===comments by dabay===Two-point lookup.‘‘‘Class Solution:# @param A, a list of integers# @param tar

Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the original List.For example,Given 1->2->3->3->4->4->5, return 1->2->5.Given 1->1->1->2->3, return 2->3.The practice is to use two pointers to the pre cur1.dummy.next= Head2.pre=dummy Cur=dummy,next3. When the pointer moves to the Pre.next and Cur.next positions, add this position to the pre.next.4. If no, skip cur.One traversal time is O (n)The code is as follows# Definition for singly-l