leetcode

map,If there is, then the request for a, a, a must come out. The index value to be evaluated is the 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 (Numbe

#-*-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

, cannot forget, if the array has no elements, return 0.4. The parameters of the max and Min functions are two values and can only be compared with two values, so the function is called two times.maxlocal = Max (max (Maxlocal*a[i], a[i]), minlocal*a[i]); minlocal = min (min (maxcopy*a[i], a[i]), minlocal*a[i]);Complexity: O (N)AC Code:Class Solution {public: int maxproduct (int a[], int n) { if (n = = 0) return 0; if (n = = 1) return a[0]; int maxlocal = a[0];

second with a null opening[2, 3] —————— the first starting with 2, at this point to 3, ending with a subset beginning with 2 , deleting 2, backtracking to an empty opening[3] —————— the last one with 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

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

. Consider s[0,1,2,..., i-1] At the end of two digits as a letter,Dp[i] "2" = IsValid (s[i-2, i-1]) = = true? 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 a

the method of calculation is the same as recursion, Zuozi (J-1) * Right subtree (i-j)3. Implementing Recursive formulasSum + = (result[j-1] * result[i-j]);4. Maintain a container of result, the number of the end of the container is the desired f (N). Therefore, each calculation must first reset 0, 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 an

int[] DP =New intN [2];4Dp[0][0] = nums[0];5 int[] max = {Dp[0][0], 0, 0};6 7 for(inti = 1; I ){8 if(Dp[i-1][0] ) {9Dp[i][0] =Nums[i];TenDP[I][1] =i; One}Else { ADp[i][0] = Nums[i] + dp[i-1][0]; -DP[I][1] = dp[i-1][1]; - } the - if(Max[0] ]) { -Max[0] = dp[i][0]; -MAX[1] = dp[i][1]; +MAX[2] =i; - } + } A at returnMax; -}Algorithm complexity: O (N)Max[] Three elements are max value, start posit

= = = ' + '? 0:strtol (VSTR1, VSTR1, ten); Long v2 = *VSTR2 = = = ' + '? 0:strtol (VSTR2, VSTR2, ten); if (V1 > V2) ret = 1; else if (V1 Answer2:c++ method idea: Use the split function directly according to the delimiter '. ' Separate the version number into the array, and then compare each. Use the Split function in the Boost library. #include STD::VectorSTD::string>STRs;Boost::Split(STRs, "string to split",Boost::is_any_of("\ T"));Code:leetco

) return a[left];4. All elements are considered well, and some values are not dropped. Add the middle to the left interval.Complexity: O (Nlog (n)) ac Code: Class Solution {public: int maxsubarray (int a[], int n) { if (n = = 0) return 0; Maxsubarray_helper (A, 0, n-1); } 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

multiple transactions at the same time1 classSolution:2 defMaxprofit (self,prices):3 """4 : 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

go.DP [M][n] is a matrix store, the min value of every location we can get.Boundary Processing , Dp[i][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 calcu

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.,

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.Idea: LRU here involves the use of vectors to record memory addre

);if(It! = M_cache.end ()) {node* cur = it->second; Cur->pre->next = cur->next; Cur->next->pre = cur->pre; Puttohead (cur);returncur->value; }Else return-1; }void Set(intKeyintValue) { mapint, node*>:: Iterator it = m_cache.find (key);if(It! = M_cache.end ()) {node* cur = it->second; Cur->value = value; Cur->pre->next = cur->next; Cur->next->pre = cur->pre; Puttohead (It->second); }Else{node* New_node

(), m_list, it); returnM_list.begin ()Val; } } void Set(intKeyintvalue) { if(M_map.find (key) = =M_map.end ()) { //Delete the back one if reach capacity if(M_capacity = =m_list.size ()) {Cachenode tmp=M_list.back (); M_list.pop_back (); M_map.erase (Tmp.key); } //Insert new one into the headcachenode node (key, value); M_list.push_front (node); M_map[key]=M_list.begin (); }

Tenth lineHow would you print just the 10th line of a file?For example, assume. has the file.txt following content:Line 1Line 2Line 3Line 4Line 5Line 6Line 7Line 8Line 9Line 10Your script should output the tenth line, which is:Line 10[Show hint]Hint:1. If The file contains less than ten lines, what should do you output?2. There ' s at least three different solutions.Try to explore all possibilities. https://leetcode.com/problems/tenth-line/ The Magic of awkawk ' nr==10 ' file. txt[

);Else returnSEARCHR (Nums, m+1, R, Target); }Else{if(Target > Nums[m] target 1])returnSEARCHR (Nums, m+1, R, Target);Else returnSEARCHR (Nums, L, M, target); } }intSearch vectorint> Nums,intTarget) {returnSEARCHR (Nums,0, Nums.size (), target); }};Array bounds [L, r]– Note the bounds of the incoming arrayclassSolution { Public://nums array boundary is [L,r] intSEARCHR ( vectorint> Nums,intLintRintTarget) {if(R return-1;intm = (l+r)/2;if(Nums[m] = = target)re