leetcode

{interval[]} intervals * @param {Interval} NewInterval * @return {interval[]}!!!!!!! wrong format*/varInsert =function(intervals, newinterval) {varres = []; varisinserted =false; for(vari = 0; i ){ 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.sta

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

-robin, each number corresponds to the index, not the 1 corresponding to the 0index,2 corresponding to 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 i

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

){ inttemp = Str.charat (i)-' 0 '; if(temp = = 0) Continue; ElseCount+= (int) Math.pow (2,len-1-i); } returncount; }}Improved thinking: Using the new generation principle, you can see the N-bit gray code consists of two parts, part of the n-1-bit gray code, plus 1Original link: Http://www.cnblogs.com/springfor/p/3889222.html?utm_source=tuicoolImproved code: Public classSolution { PublicListintN) {if(n==0) {ListNewArraylist(); Req.add (0); returnreq; } List); int

the linked list and save it in an array and build it. The direct construction method here is to use 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

-dimensional arrays are required.[CPP]View Plaincopyprint? class solution{ public :NBSP;NBSP; 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];// sele

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

left side, the left side is incremented, and the breakpoint is on the right:If target increments 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

update points to the next position of I.When the opening parenthesis ') ' is encountered,If the 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