cortana sword

@import modules in the main file.Be sure to configure the LoadPath key first, such as:loadpath:[' ' ' ],options:{Style: ' Compressed 'Compiled format: Nested is the default format (nested indentation), expanded is not indented, extended format,Compact is a compact version, compressed is compressed}，files:{' ......}}});4, there are many configuration tasks, enjoy Baidu GitHub it! JS compression merge can be used Uglify5. Load the Task plugin:Grunt.loadnpmtasks (' Grunt-contrib-watch ');Grunt.loa

intNUMBEROF1 (intN) { if(n) return0; String Number=string.valueof (n); Char[] numbers=Number.tochararray (); returnNumberof1betweenn (numbers,0); } Public intNumberof1betweenn (Char[] numbers,inti) {if(numbers==NULL|| i>numbers.length| | I) return0; //the first number processed intfirst=numbers[i]-' 0 '; intFirstnumber=0; //number of digits processed intlen=numbers.length-i; //when there's only one number if(len==1first==0) return

) { int count =0; while(n!=0){ count++; n = n(n-1); } return count; }}Analytical: If an integer is not 0, then at least one of the integers is 1. If we subtract this integer by 1, then the 1 at the far right of the integer will become 0, and all 0 after 1 will be 1 (if the rightmost 1 is followed by 0). All remaining bits will not be affected. As an example:① binary Number 1100② minus 1, get 1011.③11001011=1000Comparing ① and ③ you will find

Title Description A frog can jump up to 1 steps at a time, or jump up to level 2 ... It can also jump on n levels. To ask the frog to jump on an n-level stair total number of hops Enter a description Number of steps Output description Jump Method Number Problem analysis Set N-order hop number to F (n) When N=1, f (1) = 1 When the n=2, divided into the last step 2 and jump 1 order two cases, there is f (2) =f (0) +f (1) =1+1=2 When n=

course I was going to save it with a static array, and the next 2000 was enough, because int was super big.Class Solution {public: int arr[2005]; int getuglynumber_solution (int index) { if (indexThis can be too, but not advocated;Finally I want to say, originally I want to use the dynamic array vector, but did not succeed, subscript out of bounds, I do not know how to change the good.Can someone help me with this mistake?Reply me!Class Solution {public: vector Copyright NOTICE

sequences continues to increase until the index number is found. class solution { Public:intGetuglynumber_solution (int Index) {if(Index==1)return 1;int* x =New int[Index]; x[0] =1;intW2 =0, W3 =0, W5 =0; for(inti =1, j =1; I Index; i++) {intTMP2 = x[w2] *2;intTmp3 = x[w3] *3;intTMP5 = x[w5] *5;intt = min (tmp2, min (Tmp3, TMP5));if(t = = tmp2) w2++;if(t = = Tmp3) w3++;if(t = = tmp5) w5++; X[i] = t; }returnx[Index-1]; }}; Copyright NOTICE: This article for Bo Master original article, wit

), and the time required for each adjustment of the heap is O (logk). And a total of adjustments to n-k times. So the complexity of time isO (k + (N-K) logk), when K is far less than N, the time complexity can be approximated to O (NLOGK). In addition, the algorithm is very suitable for mass data processing, especially in memory limited. You cannot read all the data at once. When n is very large, and K is smaller, the memory is read into the K data at a time. Then each time can read into a compa

seconds may go into a trap, so early rescue small easy. 6 Input Description:7 The first behavior is an integer n (n≤1000), which indicates that the hyper-super has a total of n traps. 8 the second line has n integer xi, which indicates the horizontal axis of the first trap9 The third line has n integer yi, which indicates the ordinate of the first trapTen ensure that the coordinates are within the grassland range. One A Output Description: - output An integer that indicates that the smallest

[])Stack.peek ();int row = temp[0];int col = temp[1];if (Data[row-1][col] = = Str[point] flag[row-1][col] = =True) {Int[] Position = {row-1, col};flag[position[0]][position[1]] =FalseStack.push (position);p oint++;}Elseif (Data[row][col +1] = = Str[point] Flag[row][col +1] = =True) {Int[] Position = {row, col +1};flag[position[0]][position[1]] =FalseStack.push (position);p oint++;}Elseif (Data[row +1][col] = = Str[point] Flag[row +1][col] = =True) {Int[] Position = {row +1, col};flag[position

(1) is used to get the value of the median. Programme VI:take a big top pile and a small top heap. Although the AVL tree is more efficient, most programming language libraries do not implement this data structure, and it is a bit difficult to write their implementation code in a short time while modifying the balance factor. Here we use two containers to achieve the same effect as the program five. Here's how:(1) With two heaps, a large top heap, a small top heap. Divide the data into two parts

The title description enters a linked list, outputting the penultimate K node in the list. Train of Thought: Method 1: Set 2 pointer p,q.　　　　P moves the K-time first, and then the PQ moves back, p to the end of the list, Q points to the last K-node. Attention://need to consider k=0, and the case of list is emptyMethod 2: Use the stack, first all stack, pop k times, get the countdown k node AC code:1 /*2 struct ListNode {3 int val;4 struct ListNode *next;5 listnode (int x):6 val (x), Next (NULL)

, we can get there. CodeImport Java.util.Stack;PublicClass Solution {PublicIntMovingcount(int threshold,int rows,int cols) {int result =0;if (Threshold 0 | | Rows 0 | | Cols 0) {return result; }int[][] data =NewInt[rows +2][cols +2];for (int i =0; I 2; i++) {for (Int J =0; J 2; J + +) {if (i = =0 | | i = = (rows +1) | | j = =0 | | j = = (cols +1)) {Data[i][j] =-1; }else {Data[i][j] =0;int Rowvar = i-1;int Colvar = J-1;while (Rowvar >0) {Data[i][j] + = (rowvar%10); Rowvar/=10; }while (Colvar

, ASCII as the key value of the hash table, the position of the character corresponding to, also can be understood as subscript as the value of the hash table. For example, the array is initialized to-1, the first character that appears is 0, the second one appears is 1 ... Once again, if it appears again, set to-2, and at the end of the search, look for the first number equal to 0.The following code is primarily used to simulate character stream manipulation using strings.The specific implement

false; }}} Solution Two:By recursion, the multiplication can be reduced by doubling the multiplier each time, but it is important to check the input before using recursion.Public doublePower01all (double base,int exponent) {if (Equal (base,0.0) Exponent 0) {Return0.0;}int absexp = exponent;if (Exponent 0) {absexp =-exponent;} Doubleresult =Power01 (base, absexp);if (Exponent 0) {result =1/Result;}return result;} Public double power01 (double base, int exponent) {if (exponent = = 0

Middle-Number in Data Flow topic descriptionHow can I get the median in a data stream? If you read an odd number of values from the data stream, the median is the value in the middle after all the values have been sorted. If you read an even number of values from the data stream, the median is the average of the median two numbers after all the values are sorted.Ideas We can use arrays, we need to sort them each time we take the median We can also use a balanced binary tree, but the

;//Step-Count device - intLength=N; + while(length>0){ -indexinnums++; + //1 A if(indexinnums==N) atIndexinnums=0; - //2 - if(Nums[indexinnums]==-1) - Continue; -count++;//pedometer + +; - //3 in if(count==L) { -Nums[indexinnums]=-1; toCount=0; +length--; - } the } * returnindexinnums; $ Panax Notoginseng } - the + Pub

a function is used to determine whether a binary tree is symmetrical. If a binary tree is the same as its mirror, then it is symmetrical. For example, the following two tree chart (1) is a symmetrical two-fork tree, and the figure (2) (3) is not.Analysis: We know that there are three ways to traverse a tree: Before, in, and after. As the name implies, symmetry is the left and the right equal, the middle of their own equals.so we can define a symmetric traversal algorithm ourselves, such as front

title: Given a binary tree and one of the nodes, how to find the next node in the sequence of the middle order traversal? The nodes in the tree have a pointer to the parent node in addition to the two pointers pointing to the left and right sub-nodes respectively. Analysis: This has been said to be the middle sequence traversal, so we take the middle sequence traversal as an example. Because it is a two-fork tree, there are three kinds of cases;(1) If a node has a right sub-tree, then its next n

for each elementPublicvoid Clonerandom (Randomlistnode phead) {Randomlistnode temp = Phead;while (temp! =NULL) {if (temp.random! =NULL) {temp.Next.random = Temp.random.Next;}else {temp.Next.random =null;} temp = temp.Next.Next;}}Pulling out of the entire list, the linked list we copied.public randomlistnode reconnectnode (Randomlistnode phead) {RandomListNode pNode = PHead; Randomlistnode newhead = null; Randomlistnode NewNode = null; if (pnode! = null) {newhead = Pnode. Next;newnode = Newhead;

order to move backward one bit, until not equal, so that the loop, such as the stack sequence traversal completed, if the secondary stack is not empty, the pop-up sequence is not the stack's popup order.#-*-coding:utf-8-*-classSolution:defIspoporder (self, PUSHV, POPV):#Write code herestack = [] forXinchpushV:stack.append (x) whileStack andSTACK[-1] = =popv[0]: Stack.pop () popv.pop (0)returnFalseifStackElseTrueif __name__=='__main__': #into the stack: 1,2,3,4,5 #possible out-of-s