leetcode vs hackerrank

){ varCurr =Intervals[i]; if(Curr.end isinserted) {Res.push ([Curr.start, Curr.end]); }Else if(Newinterval.end isinserted) {Res.push ([Newinterval.start, Newinterval.end]); Res.push ([Curr.start, Curr.end]); Isinserted=true; }Else if(Curr.start //MergeNewinterval.start =math.min (Curr.start, Newinterval.start); Newinterval.end=Math.max (Curr.end, newinterval.end); }Else{ //NewInterval Eat Curr } } if(!isinserted) {Res.push ([Newinte

https://leetcode.com/problems/excel-sheet-column-title/Given a positive integer, return its corresponding column title as appear in an Excel sheet.For example: 1, A 2, B 3, C ... AB, AA, Z I hate the conversion of the system, silly calculation is not clear./**/varfunction(n ) {var res = ""; var codea = "A". charCodeAt (); while (n > 0) { n--; = String.fromCharCode (Codea + N) + res; = parseint (N/26); } return Res;}; [

https://leetcode.com/problems/number-of-1-bits/Number of 1 BitsWrite a function that takes an unsigned integer and returns the number of ' 1 ' bits it has (also known as the Hamming weigh T).For example, the 32-bit integer ' One ' 00000000000000000000000000001011 has a binary representation, so the function should return 3. /**/varfunction(n ) {var res = 0; for (i = n; I!== 0;) { if(i% 2 = = 1) { res+ +; } = parseint (I/2); } return Res;};　　

https://leetcode.com/problems/majority-element/Majority ElementGiven an array of size n, find the majority element. The majority element is the element, the appears more than times ⌊ n/2 ⌋ .Assume that the array was non-empty and the majority element always exist in the array. In order, the middle one is the right result. This problem has solution, the solution is very many./**/varfunction(nums ) {= nums.sort ( ); return nums[parseint (NUMS.LENGTH/2)];};[

https://leetcode.com/problems/jump-game/Jump GameGiven an array of non-negative integers, you is initially positioned at the first index of the array.Each element of the array represents your maximum jump length is at that position.Determine if you is able to reach the last index.For example:A = [2,3,1,1,4] , return true .A = [3,2,1,0,4] , return false . The greed of water and water/** * @param {number[]} Nums * @return {Boolean}*/varCanjump =function(nums) {varHasjump = Nums[0]; for(vari

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 must is Les S than Index2. Please note that your returned answers (both Index1 and INDEX2) is not zero-based.You may assume this each input would has exactly one solution.Input:numbers={2, 7, one, A, target=9Output:index1=1, index2=2Test instructions: Give the unordered array A to find the sum of

First on the topic:Given A string containing just the characters‘(‘,‘)‘,‘{‘,‘}‘,‘[‘and‘]‘, determine if the input string is valid.the brackets must close in the correct order," () " and" () []{} " are all valid But" (] " and" ([)] " are not. Write the answer yourself, test it throughpublic class Solution {public Boolean isValid (String s) {linkedlistfor (char C:s.tochararray ()){if (!stack.isempty ()){if (Stack.peek () ==40c==41| | Stack.peek () ==91c==93| | Stack.peek () ==123c==125){Stack.pop

the 1 index ...)1 is the root node, 1 of the Zuozi is 0, the right subtree is 2,3 (two). So is 1 is the number of types of root nodes: Res[0] times res[2] number of cases.When 2 is the root node, 2 of the Zuozi are 1 (1) and the right subtree is 1 (1). So is 2 is the number of types of root nodes: res[1] times res[1] number of cases.3 is the root node, 3 of the Zuozi is (2), the right subtree is (0). So is 2 is the number of types of root nodes: res[2] times res[0] number of cases.1 Public cla

Given a binary tree, find the maximum path sum.The path may start and end at any node in the tree.For example:Given the below binary tree, 1 / 2 3Return 6 .Problem Solving Ideas:DFS brute force enumeration, note that if the static global variable is used, it can be passed inside the IDE, but cannot be tested on OJ, so you need to create a class to store the results, and the Java implementation is as follows:public class Solution {static public int maxpathsum (TreeNode root) {re

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.For example,"A man, a plan, a canal: Panama"is a palindrome."race a car"is not a palindrome.Problem Solving Ideas:Note the topic, ignoring case, ignoring non-letters or numbers, Java implementation is as follows: public Boolean ispalindrome (String s) { if (s.length () Java for Leetcode Valid palindrome

Say you has an array for which the i-th element is the price of a given-stock on day I.Design an algorithm to find the maximum profit. You are in most of the transactions.Note:Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).Problem Solving Ideas:Since it is a two-time deal, it is divided into two areas. First, according to the idea of a transaction to calculate the maximum value of the transaction, in an array, and then forward from the back,

Follow up for "Remove duplicates":What if duplicates is allowed at the most twice?For example,Given sorted array nums = [1,1,1,2,2,3] ,Your function should return length =, with the first 5 five elements of nums being 1 , 1 2 2 ,, and 3 . It doesn ' t matter what are you leave beyond the new length.Problem Solving Ideas:There are many ways to do this, and here is another way to open an array, Java implementation is as follows: public int removeduplicates (int[] nums) {if (Nums.length Java for

) {String temp=Gray[i]; intToInt =Binarytoint (temp); Req.add (ToInt); } returnreq; } PublicString[] Factory (intN) {//Build Gray Codestring[] req =Newstring[(int) Math.pow (2, N)]; if(n = = 0) {req[0] = "0"; returnreq; } if(n = = 1) {req[0] = "0"; req[1] = "1"; returnreq; } intup = N/2;//Small intleft = n-up;//Bigstring[] Upstr=factory (UP); String[] Leftstr=factory (left); intLen = 0; for(inti = 0; i ){ for(intj = 0; J ) {Req[len

the fast pointer to find the midpoint of the list, the left half of the list ends point to null, and recursively constructs a left subtree. The right half of the part is built recursively into the right sub-tree. If you do not want to destroy the original linked list, then you can use the first to save to the array and then build the method.1 PublicTreeNode Sortedlisttobst (ListNode head) {2 if(head==NULL)3 return NULL;4 if(Head.next = =NULL)5 return