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  • is this code correct? [closed]

    - by davit-datuashvili
    hi i have poste this code from this title http://stackoverflow.com/questions/2896363/hi-i-have-question-here-is-pseudo-code-about-sift-up-and-sift-down-on-heaps i have following code of siftup on heap is it correct?i have put here because i have changed at old place my question and it became unreadable so i have posted here public class siftup{ public static void main(String[]args){ int p; int n=12; int a[]=new int[]{15,20,12,29,23,17,22,35,40,26,51,19}; int i=n-1; while (i!=0){ if (i==1) break; p=i/2; if (a[p]<=a[i]){ int t=a[p]; a[p]=a[i]; a[i]=t; } i=p; } for (int j=0;j<n;j++){ System.out.println(a[j]); } } } //result is this 15 20 19 29 23 12 22 35 40 26 51 17 is it correct?

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  • Given the lat/lon of 2 close points on earth (<10m), How do I calculate the distance in metres?

    - by Rory
    I have the lat/lon of 2 points on the earth. They are really close together, <10m. Let's assume the earth is flat. How do I calculate the distance between them in metres? I know about tools (PostGIS, etc.) that can do this correctly, however I'm just doing a rough and ready type, and I'm OK with low accuracy. At such small sizes a difference of 1% is only 10cm, which is fine for me. I'm doing this in stock python. I'm OK with a standard Euclidean distance thing.

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  • Doubling a number - shift left vs. multiplication

    - by ToxicAvenger
    What are the differences between int size = (int)((length * 200L) / 100L); // (1) and int size = length << 1; // (2) (length is int in both cases) I assume both code snippets want to double the length parameter. I'd be tempted to use (2) ... so are there any advantages for using (1)? I looked at the edge cases when overflow occurs, and both versions seem to have the same behavior. Please tell me what am I missing.

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  • Java How to find a value in a linked list iteratively and recursively

    - by Roxy
    Hi I have a method that has a reference to a linked list and a int value. So, this method would count and return how often the value happens in the linked list. So, I decided to make a class, public class ListNode{ public ListNode (int v, ListNode n) {value = v; next = n;) public int value; public ListNode next; } Then, the method would start with a public static int findValue(ListNode x, int valueToCount){ // so would I do it like this?? I don't know how to find the value, // like do I check it? for (int i =0; i< x.length ;i++){ valueToCount += valueToCount; } So, I CHANGED this part, If I did this recursively, then I would have public static int findValue(ListNode x, int valueToCount) { if (x.next != null && x.value == valueToCount { return 1 + findValue(x, valueToCount);} else return new findvalue(x, valueToCount); SO, is the recursive part correct now?

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  • Fastest method to define whether a number is a triangular number

    - by psihodelia
    A triangular number is the sum of the n natural numbers from 1 to n. What is the fastest method to find whether a given positive integer number is a triangular one? I suppose, there must be a hidden pattern in a binary representation of such numbers (like if you need to find whether a number is even/odd you check its least significant bit). Here is a cut of the first 1200th up to 1300th triangular numbers, you can easily see a bit-pattern here (if not, try to zoom out): (720600, '10101111111011011000') (721801, '10110000001110001001') (723003, '10110000100000111011') (724206, '10110000110011101110') (725410, '10110001000110100010') (726615, '10110001011001010111') (727821, '10110001101100001101') (729028, '10110001111111000100') (730236, '10110010010001111100') (731445, '10110010100100110101') (732655, '10110010110111101111') (733866, '10110011001010101010') (735078, '10110011011101100110') (736291, '10110011110000100011') (737505, '10110100000011100001') (738720, '10110100010110100000') (739936, '10110100101001100000') (741153, '10110100111100100001') (742371, '10110101001111100011') (743590, '10110101100010100110') (744810, '10110101110101101010') (746031, '10110110001000101111') (747253, '10110110011011110101') (748476, '10110110101110111100') (749700, '10110111000010000100') (750925, '10110111010101001101') (752151, '10110111101000010111') (753378, '10110111111011100010') (754606, '10111000001110101110') (755835, '10111000100001111011') (757065, '10111000110101001001') (758296, '10111001001000011000') (759528, '10111001011011101000') (760761, '10111001101110111001') (761995, '10111010000010001011') (763230, '10111010010101011110') (764466, '10111010101000110010') (765703, '10111010111100000111') (766941, '10111011001111011101') (768180, '10111011100010110100') (769420, '10111011110110001100') (770661, '10111100001001100101') (771903, '10111100011100111111') (773146, '10111100110000011010') (774390, '10111101000011110110') (775635, '10111101010111010011') (776881, '10111101101010110001') (778128, '10111101111110010000') (779376, '10111110010001110000') (780625, '10111110100101010001') (781875, '10111110111000110011') (783126, '10111111001100010110') (784378, '10111111011111111010') (785631, '10111111110011011111') (786885, '11000000000111000101') (788140, '11000000011010101100') (789396, '11000000101110010100') (790653, '11000001000001111101') (791911, '11000001010101100111') (793170, '11000001101001010010') (794430, '11000001111100111110') (795691, '11000010010000101011') (796953, '11000010100100011001') (798216, '11000010111000001000') (799480, '11000011001011111000') (800745, '11000011011111101001') (802011, '11000011110011011011') (803278, '11000100000111001110') (804546, '11000100011011000010') (805815, '11000100101110110111') (807085, '11000101000010101101') (808356, '11000101010110100100') (809628, '11000101101010011100') (810901, '11000101111110010101') (812175, '11000110010010001111') (813450, '11000110100110001010') (814726, '11000110111010000110') (816003, '11000111001110000011') (817281, '11000111100010000001') (818560, '11000111110110000000') (819840, '11001000001010000000') (821121, '11001000011110000001') (822403, '11001000110010000011') (823686, '11001001000110000110') (824970, '11001001011010001010') (826255, '11001001101110001111') (827541, '11001010000010010101') (828828, '11001010010110011100') (830116, '11001010101010100100') (831405, '11001010111110101101') (832695, '11001011010010110111') (833986, '11001011100111000010') (835278, '11001011111011001110') (836571, '11001100001111011011') (837865, '11001100100011101001') (839160, '11001100110111111000') (840456, '11001101001100001000') (841753, '11001101100000011001') (843051, '11001101110100101011') (844350, '11001110001000111110') For example, can you also see a rotated normal distribution curve, represented by zeros between 807085 and 831405?

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  • How to figure out "progress" while sorting?

    - by Mehrdad
    I'm using stable_sort to sort a large vector. The sorting takes on the order of a few seconds (say, 5-10 seconds), and I would like to display a progress bar to the user showing how much of the sorting is done so far. But (even if I was to write my own sorting routine) how can I tell how much progress I have made, and how much more there is left to go? I don't need it to be exact, but I need it to be "reasonable" (i.e. reasonably linear, not faked, and certainly not backtracking).

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  • How to calculate the state of a graph?

    - by zcb
    Given a graph G=(V,E), each node i is associated with 'Ci' number of objects. At each step, for every node i, the Ci objects will be taken away by the neighbors of i equally. After K steps, output the number of objects of the top five nodes which has the most objects. Some Constrains: |V|<10^5, |E|<2*10^5, K<10^7, Ci<1000 My current idea is: represent the transformation in each step with a matrix. This problem is converted to the calculation of the power of matrix. But this solution is much too slow considering |V| can be 10^5. Is there any faster way to do it?

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  • Weakly connected balanced digraph

    - by user1074557
    How can I prove that if a balanced digraph is weakly connected, then it is also strongly connected? (balanced digraph means that for every node, it's indegree and outdegree is the same and weakly connected means the non-directed version of this graph is connected). What I can think of so far is: if the graph is balanced, it means it is a union of directed cycles. So if I remove any cycle, it will stay balanced. Also each vertex in the cycle has one edge coming into it and one edge leading out of it.. Then I guess I need to use some contradiction or induction to prove that the graph is strongly connected.. That's where I confused.

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  • determine if line segment is inside polygon

    - by dato
    suppose we have simple polygon(without holes) with vertices (v0,v1,....vn) my aim is to determine if for given point p(x,y) any line segment connecting this point and any vertices of polygon is inside polygon or even for given two point p(x0,y0) `p(x1,y1)` line segment connecting these two point is inside polygon? i have searched many sites about this ,but i am still confused,generally i think we have to compare coordinates of vertices and by determing coordinates of which point is less or greater to another point's coordinates,we could determine location of any line segment,but i am not sure how correct is this,please help me

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  • C# compare algorithms

    - by public static
    Hi, Are there any open source algorithms in c# that solve the problem of creating a difference between two text files? It would be super cool if it had some way of highlighting what exact areas where changed in the text document also.

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  • Given a 2d array sorted in increasing order from left to right and top to bottom, what is the best w

    - by Phukab
    I was recently given this interview question and I'm curious what a good solution to it would be. Say I'm given a 2d array where all the numbers in the array are in increasing order from left to right and top to bottom. What is the best way to search and determine if a target number is in the array? Now, my first inclination is to utilize a binary search since my data is sorted. I can determine if a number is in a single row in O(log N) time. However, it is the 2 directions that throw me off. Another solution I could use, if I could be sure the matrix is n x n, is to start at the middle. If the middle value is less than my target, then I can be sure it is in the left square portion of the matrix from the middle. I then move diagnally and check again, reducing the size of the square that the target could potentially be in until I have honed in on the target number. Does anyone have any good ideas on solving this problem? Example array: Sorted left to right, top to bottom. 1 2 4 5 6 2 3 5 7 8 4 6 8 9 10 5 8 9 10 11

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  • generate k distinct number less then n

    - by davit-datuashvili
    hi i have following question task is this generate k distinct positive numbers less then n without duplication my method is following first create array size of k where we should write these numbers int a[]=new int[k]; //now i am going to cretae another array where i check if (at given number position is 1 then generate number again else put this number in a array and continue cycle i put here a piece of code and explanations int a[]=new int[k]; int t[]=new int[n+1]; Random r=new Random(); for (int i==0;i<t.length;i++){ t[i]=0;//initialize it to zero } int m=0;//initialize it also for (int i=0;i<a.length;i++){ m=r.nextInt(n);//random element between 0 and n if (t[m]==1){ //i have problem with this i want in case of duplication element occurs repeats this steps afain until there will be different number else{ t[m]=1; x[i]=m; } } so i fill concret my problem if t[m]==1 it means that this element occurs already so i want to generate new number but problem is that number of generated numbers will not be k beacuse if i==0 and occurs duplicate element and we write continue then it will switch at i==1 i need like goto for repeat step or for (int i=0;i<x.length;i++){ loop: m=r.nextInt(n); if ( x[m]==1){ continue loop; } else{ x[m]=1; a[i]=m; continue;//continue next step at i=1 and so on } } i need this code in java please help

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  • How to find the largest power of 2 less than the given number

    - by nazar_art
    I need to find the largest power of 2 less than the given number. And I stuck and can't find any solution. Code: public class MathPow { public int largestPowerOf2 (int n) { int res = 2; while (res < n) { res =(int)Math.pow(res, 2); } return res; } } This doesn't work correctly. Testing output: Arguments Actual Expected ------------------------- 9 16 8 100 256 64 1000 65536 512 64 256 32 How to solve this issue?

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  • Algorithms after load-balancer?

    - by Vimvq1987
    I need to study about load-balancers, such as Network Load Balancing, Linux Virtual Server, HAProxy,...There're somethings under-the-hood I need to know: What algorithms/technologies are used in these load-balancers? Which is the most popular? most effective? I expect that these algorithms/technologies will not be too complicated. Are there some resources written about them? Thank you very much for your help.

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  • Special simple random number generator

    - by psihodelia
    How to create a function, which on every call generates a random integer number? This number must be most random as possible (according to uniform distribution). It is only allowed to use one static variable and at most 3 elementary steps, where each step consists of only one basic arithmetic operation of arity 1 or 2. Example: int myrandom(void){ static int x; x = some_step1; x = some_step2; x = some_step3; return x; } Basic arithmetic operations are +,-,%,and, not, xor, or, left shift, right shift, multiplication and division. Of course, no rand(), random() or similar staff is allowed.

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  • How to implement Pentago AI algorithm

    - by itsho
    Hi, i'm trying to develop Pentago-game in c#. right now i'm having 2 players mode which working just fine. the problem is, that i want One player mode (against computer), but unfortunately, all implements of minimax / negamax are for one step calculated. butin Pentago, every player need to do two things (place marble, and rotate one of the inner-boards) I didn't figure out how to implement both rotate part & placing the marble, and i would love someone to guide me with this. if you're not familiar with the game, here's a link to the game. if anyone want's, i can upload my code somewhere if that's relevant. thank you very much in advance

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  • i have question about job

    - by davit-datuashvili
    hi i will explain my situation i am study algorithms and data structures and i know many thing practical and others most thoereticaly and trying to implement in practise i want to get job where i will study more quickly i am interested if i have chance to start job when i have not expierence? thanks please advise what to do

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  • Java: Efficient Equivalent to Removing while Iterating a Collection

    - by Claudiu
    Hello everyone. We all know you can't do this: for (Object i : l) if (condition(i)) l.remove(i); ConcurrentModificationException etc... this apparently works sometimes, but not always. Here's some specific code: public static void main(String[] args) { Collection<Integer> l = new ArrayList<Integer>(); for (int i=0; i < 10; ++i) { l.add(new Integer(4)); l.add(new Integer(5)); l.add(new Integer(6)); } for (Integer i : l) { if (i.intValue() == 5) l.remove(i); } System.out.println(l); } This, of course, results in: Exception in thread "main" java.util.ConcurrentModificationException ...even though multiple threads aren't doing it... Anyway. What's the best solution to this problem? "Best" here means most time and space efficient (I realize you can't always have both!) I'm also using an arbitrary Collection here, not necessarily an ArrayList, so you can't rely on get.

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  • longest increasing subsequent

    - by davit-datuashvili
    i have write this code is it correct? public class subsequent{ public static void main(String[] args){ int a[]=new int[]{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int a_b[]=new int[a.length]; a_b[0]=1; int int_max=0; int int_index=0; for (int i=0;i<a.length;i++){ for (int j=0;j<i;j++){ if (a[i]>a[j] && a_b[i]<(a_b[j]+1)){ a_b[i]=a_b[j]+1; } } if (a_b[i]>int_max){ int_max=a_b[i]; int_index=i; } } int k=int_max+1; int list[]=new int[k]; for (int i=int_index;i>0;i--){ if (a_b[i]==k-1){ list[k-1]=a[i]; k=a_b[i]; } } for (int g=0;g<list.length;g++){ System.out.println(list[g]); } } }

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