leetcode vs hackerrank

A robot is located at the Top-left corner of a m x N grid (marked ' Start ' in the diagram below).The robot can only move either down or right at any point in time. The robot is trying-to-reach the bottom-right corner of the grid (marked ' Finish ' in the diagram below).How many possible unique paths is there?Above is a 3 x 7 grid. How many possible unique paths is there?There are two ways of solving this problem.1. Direct mathematical expression Choose M-1 from M+n-2Class solution: # @return

Title: Given a string, the length of the oldest string with no repeating character is obtained. Idea: For loop once, time complexity is O (N). The ASCII value of the character is 32~126. Start represents the initial position of the current non-repeating character string, with an initial value of 0, and a positional array of pos[128] that represents the position of the corresponding character at the current position of the for Loop index. If the current character is S[i] (its ASCII value is Cur

returnhead;4ListNode Fakehead =NewListNode (-1);5Fakehead.next =head;6ListNode ptr1 = head, PTR2 = fakehead.next.next, Newstart =Fakehead;7 intLen = 0;8 while(Ptr1! =NULL) {9len++;TenPTR1 =Ptr1.next; One } APTR1 =Fakehead.next; - if(n!=len) -n = n%Len; the intK = len/N; - for(intj = 0; J ) { - - for(inti = 1; I ) { +Ptr1.next =Ptr2.next; -Ptr2.next =Newstart.next; +Newstart.next =ptr2; APTR2 =Ptr1.next; at }

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '26:remove Duplicates from Sorted Arrayhttps://oj.leetcode.com/problems/remove-duplicates-from-sorted-array/Given a sorted array, remove the duplicates in place such, all element appear only once and return the new length.Do the allocate extra space for another array, and you must does this on place with constant memory.For example,Given input Array A = [1,1,2],Your function should return length = 2, and A is now [all].===comments by dabay===

whether the letters in the four direction match word[index+1], the match will enter the next level of recursion, so the iteration termination condition is if index reaches Word.size ()-1. Indicates that word characters are matched successfully and returns true. If you select index = = Word.size () to terminate, when index = Word.size ()-1, the next call to Word[index+1] will cross the pointer. If all characters of Word match successfully, index reaches Word.size ()-1, recursion ends, returns tr

index value of a, bBoth of these methods ensure that the INDEX1Solution Java Code:Importjava.util.Arrays;ImportJava.util.HashMap;ImportJava.util.Map; Public classTwosum { Public Static voidMain (string[] args) {int[] numbers={2, 7, 11, 15}; inttarget = 9; int[] Twosum =twosum (Numbers,target); System.out.println ("The sum indices is" + twosum[0] + "and" + twosum[1]); int[] twoSum2 =twoSum2 (Numbers,target); System.out.println ("The sum indices is" + twosum2[0] + "and" + twosum2[1]); }Idea 1 Pub

#-*-Coding:utf8-*-‘‘‘https://oj.leetcode.com/problems/container-with-most-water/Given n non-negative integers a1, A2, ..., an, where each represents a point at coordinate (I, AI).n vertical lines is drawn such, the and the other endpoints of line I am at (i, AI) and (i, 0).Find lines, which together with X-axis forms a container, such that the container contains the most water.Note:you may not slant the container.===comments by dabay===The problem is classic, but I still don't know why I can. Ju

specific problems, careful analysis and thinking, thoroughly understand the method, can do to raise one and anti-three. Attention:1. Note that the multiplication is considered in this problem, and it is necessary to consider the characteristics of multiplication and to maintain a variable minlocal.2. Note that when we are dynamically planning, the last maxlocal[i] is used when solving minlocal[i+1], so we need to save the maxlocal[i before the local optimal update]int maxcopy = maxlocal;

an empty openingextension : What should I do when nums contains repeating characters ?1, the beginning is disorderly order, so need to nums sorting to determine whether to repeat.2. In the loop of backtracking method, the first line is added to repeat judgment, if the current element (not the first) and the previous element, then skip this element does not use recursion.1 PublicListint[] nums) {2ListNewArraylist();3 Arrays.sort (nums);4Backtrack (list,NewArraylist);5 returnlist;6 }7 8 Priv

Pascal ' s Triangle:Given numrows, generate the first numrows of Pascal ' s triangle.For example, given numrows = 5,Return[ 1], [ 1,2,1], [ 1,3,3,1], [ 1,4,6,4,1]The number of rows is known to generate the Pascal Triangle. In fact, as long as there is layer I, then you can generate the first i+1 layer. Each newly generated layer is added to the final collection.1 PublicListintnumrows) {2listNewArraylist();3 4 for(inti=0;i) {5listNewArraylist();6 for(i

Title Description: According to the provisions, the string decoding, specific examples see the topic link Idea: Use two stacks to store numbers and letters, respectively Note 1: Numbers are multi-bit words, to be processed after entering the digital stack Note 2: The combined string generated during the stack will continue into the letter stack Note 3: Remember that when the alphabet is out of the stack, the characters are combined into strings. Note 4: The reason for using t

Topic:Follow up for "Remove duplicates":What if duplicates is allowed at the most twice?For example,Given sorted Array A = [1,1,1,2,2,3] ,Your function should return length = 5 , and A is now [1,1,2,2,3] .Ideas:You can draw on the two pointer movement method of the remove duplicates from Sorted array.However, some details need to be considered. For example, we need to continue to assign values that are less than 3 repetitions to the past, and more than 3 to move only the point pointer, searching

? Dp[i-2]: 03. Dp[i] = Dp[i] "1" + Dp[i] "2" Attention:1. The setting of the initial conditions dp[0], need to be based on the specific meaning of the code to determine, to meet the actual meaning (dp[2] + = Dp[0], here dp[0] means s[0,1] for a letter, so take 1) .2. If there is a mismatch of 0 in the sequence, then the decoding method is 0. This is easy to overlook, so be sure to pay attention. If a character is 0, it cannot be decoded, only considering whether it can make a valid character

, the iteration to obtain the results and then push into the container.for (int i = 2; I {int sum = 0;Calculates the number of binary lookup trees that contain I nodes, and J indicates that 1~i takes any node as the root.for (int j = 1; J {Sum + = (result[j-1] * result[i-j]); Left Tree * Right tree implementation recursive formula}result.push_back (sum);}complexity: The number of two forks to find a tree for each of the I nodes needs an i-step loop, which requires n times in total. Total time co

than 0, then directly add, otherwise take nums[i] is the largest)"Notice is not to say that dp[i] is the maximum value that a subsequence can achieve in a sequence ending with nums[i]"So the largest value in the dp[] is the largest one required.Optimization:As long as you know the maximum value of DP, you do not need to create a DP, just use a variable mem to record the current, and then see if it is larger than Max:1 Public intMaxsubarray (int[] nums) {2 intmax = Nums[0];3

) AC Code: Class Solution {public: int compareversion (string version1, String version2) { int ret = 0; Converts a string into a C-style character array char* vStr1 = (char*) version1.c_str (); char* vStr2 = (char*) version2.c_str (); while (ret = = 0 (*vstr1! = ' *vstr2 ') { //strtol will scan the parameter nptr string, skipping the preceding space character, The conversion is not started until a number

); } int maxsubarray_helper (int a[], int left, int. right) { if (left = right) return a[left]; int middle = left + (right-left)/2; Calculates the left and right interval int leftans = Maxsubarray_helper (A, Ieft, middle); int Rightans = Maxsubarray_helper (A, middle+1, right); Calculates the interval int leftmax = A[middle] with intermediate elements; int rightmax = a[middle+1]; int tmp = 0;

: Type Prices:list[int]5 : Rtype:int6 """7sum =08 forIinchRange (1, Len (Prices)):9A = Prices[i]-prices[i-1]Ten ifA >0: Onesum = sum +a A returnSumView Code3. Rotating an arrayGiven an array, move the elements in the array to the right by the K position, where K is a non-negative1 class Solution: 2 def Rotate (self,nums,k): 3 i =04 while I K:5 nums.insert (0, Nums.pop ())6 i +=1View Code4. Duplication ex

][j] represents the smallest path and from the upper-left corner to Grid[i-1][j-1]. DP[RI][CI] = grid[ri-1][ci-1] + min (Dp[ri-1][ci], dp[ri][ci-1]);AC Code:Class Solution {public: int minpathsum (vectorOptimization algorithm: Reduced space complexity to O (min (row, col))Idea:Dp[i][j] about only the previous row of Dp[i-1][j] and the previous column of Dp[i][j-1], Dp[i][j] is a line of calculation (according to the Loop), Dp[j-1] represents the value of the previous column (has been updated

Design and implement a data structure for Least frequently Used (LFU) cache. It should support the following operations: get and put .get(key)-Get The value ('ll always be positive) of the key if the key exists in the cache, otherwise return-1.put(key, value)-Set or insert the value if the key is not already present. When the cache is reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e.,