leetcode

Sort a linked list in O(n log n) time using constant space complexity.Common sorting methods are many, insert sort, choose Sort, heap sort, quick sort, bubble sort, merge sort, bucket sort, etc... Their time complexity is not the same, and here the topic limit time must be O (NLGN), meet the requirements of only fast sorting, merge sort, heap sorting, and according to the characteristics of single-linked list, the most suitable for merging sort. The code is as follows:/** Definition for singly-l

Follow up to problem "populating Next right pointers in each Node".What if the given tree could is any binary tree? Would your previous solution still work?Note:Constant extra space.For example,Given The following binary tree,1/ \2 3/ \ \4 5 7After calling your function, the tree is should look like:1, NULL/ \2, 3, NULL/ \ \4-> 5-7, NULLThe difference between this question and the original topic is that the tree here is not full tree. or a deformity. So the best way is to level order travesral

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '20:valid parentheseshttps://oj.leetcode.com/problems/valid-parentheses/Given A string containing just the characters ' (', ') ', ' {', '} ', ' [' and '] ', determine if the input string is valid.The brackets must close in the correct order, "()" and "() []{}" is all valid but "(]" and "([)]" is not.===comments by dabay===Use a stack to save the opening parenthesis, and if it comes with a corresponding closing parenthesis, pop one, or return

- while(1) - { thenow = now+K; - -temp = nums[now%numssize];//Swap once -Nums[now%numssize] =Data_to_next; +Data_to_next =temp; -cnt--; + A if((now%numssize) = = start)//if the switch to start starts, it ends this jump exchange at Break; - - } - - if(cnt==0) - Break; in } - to +}Python:1 classSolution:2 defrotate (self, nums,

Tag: false char AC Ali Leetcode [] character if else checktitle : valid parentheses sequenceDifficulty : Easytopic content :Given A string containing just the characters ‘(‘ ,,, ‘)‘ , and ‘{‘ ‘}‘ ‘[‘ ‘]‘ , determine if the input string is valid.An input string is valid if: Open brackets must is closed by the same type of brackets. Open brackets must is closed in the correct order. Note that a empty string is also considered valid.tra

it =Cachemap[key]; - Cachelist.splice (Cachelist.begin (), cachelist, it); -Cachemap[key] =Cachelist.begin (); - returnCachelist.begin ()Val; +}Else - return-1; + } A at void Set(intKeyintvalue) { - if(Cachemap.find (key)! = Cachemap.end ()) {//the element exists in the list -Auto it =Cachemap[key]; - Cachelist.splice (Cachelist.begin (), cachelist, it); -Cachemap[key] = =Cachelist.begin (); -Cachelist.begin ()->val =value; in}Else{//The element doe

Tenth line:https://leetcode.com/problems/tenth-line/Test Instructions: prints the 10th line of a file.The problem involves several points:1. How do I read a file?2. How do I find line 10th?3. What if the file is empty or less than 10 lines?Solution:1.shell a way to read a file:While read Linedo echo $linedone 2. Use the Count method to find the 10th line. 3. If the file is empty or less than 10 lines, the output is empty.Code:Cnt=1while Read Linedo if [$ (cnt)] = = [ten]; then echo

{public:lrucache () {} LRUCache (int capacity): capacity (capacity) {} int get (int key) { if (cache.find (key) = = Cache.end ()) return-1; Items.splice (Items.begin (), items, Cache[key]); Update the location of the access node and place it on the list head return cache[key]->second; } void set (int key, int value) {if (Cache.find (key) = = Cache.end ()) {///If the value to be inserted is not in the cache, the Insert it into the list header and update the iterator in the cache if (c

Design and implement a data structure for Least recently Used (LRU) cache. It should support the following operations: get and set .get(key)-Get The value ('ll always be positive) of the key if the key exists in the cache, otherwise return-1.set(key, value)-Set or insert the value if the key is not already present. When the cache is reached its capacity, it should invalidate the least recently used item before inserting a new item.Import Java.util.hashmap;import Java.util.linkedlist;class listno

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '55:jump Gamehttps://leetcode.com/problems/jump-game/Given 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.= = = Comments by dabay===Each calculation can reach a new range.If the new range goes bey

(NULL) {}7 * };8 */9 classSolution {Ten Public: Onelistnode* Deleteduplicates (listnode*head) { A if(head = = NULL)returnNULL; -ListNode *left =head; - while(left left->next left->val = = left->next->val) { the intval = left->Val; - while(left Left->val = =val) -left = left->Next; - } + if(left = = NULL)returnNULL; -Head =Left ; +ListNode *Right ; A while(left) { atright = Left->Next; - while(Right right->nex

Given an array of integers, determine if there are duplicate elements.If any value occurs at least two times in the array, the function returns True. Returns False if each element in the array is not the same.C++classSolution { Public: BOOLContainsduplicate (vectorint>nums) {Unordered_mapint,int>m; for(inti =0; I ) { if(M.find (nums[i])! =m.end ())return true; ++M[nums[i]]; } return false; }};C bool containsduplicate (int * nums, Span style= "COLOR: #0000ff" >int Numssi

#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '53:maximum Subarrayhttps://leetcode.com/problems/maximum-subarray/Find the contiguous subarray within an array (containing at least one number) which have the largest sum.For example, given the array [−2,1,−3,4,−1,2,1,−5,4],The contiguous subarray [4,−1,2,1] has the largest sum = 6.More Practice:If you had figured out the O (n) solution, try coding another solution using the divide and conquer approach,Which is more subtle.= = = Comments by

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. Transactions at the most K .Note:Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).Credits:Special thanks to @Freezen for adding this problem and creating all test cases.Based on Dynamic PlanningMaintain II vectors:gpro:to Day I the maximum profit lpro:to Day I, the maximum profit with jth sell by day IThe complex

Given an input string, reverse the string word by word.For example,Given s = " the sky is blue ",Return " blue is sky the ".Update (2015-02-12):For C programmers:try to solve it in-place in O(1) space.Linear time:1. Go through the string detect whether the "if not append the char to Word2 If the place is ' and word was not ', append the word to res3. Go to the end append the word if the word was not ""Reverse the Res4. Do join.Class solution: # @param s, a string # @return A string def

result, the following iteration formula is available:LOCAL_MAX[I][J] = max (local_max[i-1][j]+diff, Global_max[i-1][j-1]+max (diff,0))The above recursive formula regards Local_max as the goal of optimization, and the idea is to fellow the classical dynamic planning.However, there is a part I have not figured out (blue word part), according to the classic dynamic planning ideas intuitive to analyze, it should be local_max[i-1][j] ah, how to come out a diff it?

(n); for(inti = 1; I ) { if(i > N) Break; DFS (i, res, n); } returnRes; } Private voidDfsintI, listintN) {res.add (i); for(intm = 0; M ) { if(I * ten + M > N) Break; DFS (i* 10 +m, res, n); } }}Solution Two (Java)classSolution { PublicListintN) {ListNewArraylist(n); Res.add (1); intPre = 1; for(inti = 1; I ) { if(PRE * ; Else { while(Pre% = = 9 | | pre = n) Pre/= 10; Pre++; } res.add (pre); } retu

# #002 Input: (2, 4, 3) + (5, 6, 4) output:7, 0, 8# linked list nodes are a single number, which can be considered 243+564=807#先定义节点和链表类Import NumPy as NPImport timeClass Node (object):def __init__ (Self,n,next_node=none):Self.data=nSelf.next=next_nodeClass Linklist (object):def __init__ (self):Self.head=nonedef init (self,data):Assert type (data) ==list,type (data)Self.head=node (Data[0],none)P=self.headFor i in Data[1:]:Node=node (i)P.next=nodeP=p.nextDef show (self):L=[]P=self.headWhile P:L.a

continue looking for a new interval.Java/*** Definition for a interval. * public class Interval {* int start; * int end; * Interval () {start = 0; end = 0; } * Interval (int s, int e) {start = s; end = e;} }*/classSolution { PublicListintervals) { intLen =intervals.size (); if(Len returnintervals; Collections.sort (intervals,NewCMP ()); ListNewArraylist(); intStart = Intervals.get (0). Start, end = Intervals.get (0). End; for(inti = 1; i ) { if(Intervals.get (i). Start end) {

Given words (beginword and Endword), and a dictionary ' s word list, find the length of shortest trans Formation sequence from Beginword to Endword, such that: Only one letter can is changed at a time. Each transformed word must exist in the word list. Note that Beginword was not a transformed word. Note: Return 0 If there is no such transformation sequence. All words has the same length. All words contain only lowercase alphabetic characters. Assume no du