algorithm

问题 Since Dijkstra's algorithm and Prim's algorithm are so similar, I'd like to make a base algorithm class (I'll call it "Greedy" or something) and then I want to inherit from Greedy and tweak the algorithm based on the class. I think it boils down to this. I'd like to reuse a large portion of an algorithm but tweak an operation or two. class BaseAlg { public: BaseAlg(std::vector<int> data) //constructor sums a vector and stores result { int accum = 0; for (unsigned int i = 0; i < data.size(); +

问题 I'm using Heap's algorithm to create a list-of-lists containing each permutation of said list. Each permutation will be its own list. It works properly when I print it within the algorithm, but it doesn't work properly when I try to add it to my list-of-lists and they are all the same array (4, 1, 2, 3). I commented out the print that I tested to make sure it was working. My current code: public static ArrayList<int[]> lists = new ArrayList<>(); public static void main(String[] args) { int[]

问题 I am looking for a distributed semaphore implementation (with postgres / zookeeper as store) that is similar to the concept of java.util.concurrent.Semaphore which will maintain a set of permits that will be taken using acquire() and released with release() allowing me to restrict access to some resource or synchronize some execution. The only difference is that this semaphore should allow me to do all these actions across multiple jvms. Can someone point to me if there is any such

问题 This question already has answers here : How do you split a list into evenly sized chunks? (63 answers) Closed 6 years ago . Is there an easy way to convert a 1D list into a 2D list with a given row length? Suppose I have a list like this: myList = [1, 2, 3, 4, 5, 6, 7, 8, 9] I want to convert the above list into a 3 x 3 table like below: myList = [[1,2,3],[4,5,6],[7,8,9]] I know I can accomplish this by creating a new 2D list called myList2 and inserting the element into MyList2 using 2

问题 I know of an algorithm (it can be found online) to rank a permutation, i.e. given a permutation return the integer index into the list of lexicographically-sorted permutations, but I don't know any unrank algorithm that does the opposite: given an index i, return the i-th permutation in that lexicographic order. Since I couldn't find any, can somebody please shed some light? 回答1: Let's say you are permutating the letters (a, b, c). There are 3×2×1=6 permutations. Out of these, a third starts

问题 I know of an algorithm (it can be found online) to rank a permutation, i.e. given a permutation return the integer index into the list of lexicographically-sorted permutations, but I don't know any unrank algorithm that does the opposite: given an index i, return the i-th permutation in that lexicographic order. Since I couldn't find any, can somebody please shed some light? 回答1: Let's say you are permutating the letters (a, b, c). There are 3×2×1=6 permutations. Out of these, a third starts

问题 I know of an algorithm (it can be found online) to rank a permutation, i.e. given a permutation return the integer index into the list of lexicographically-sorted permutations, but I don't know any unrank algorithm that does the opposite: given an index i, return the i-th permutation in that lexicographic order. Since I couldn't find any, can somebody please shed some light? 回答1: Let's say you are permutating the letters (a, b, c). There are 3×2×1=6 permutations. Out of these, a third starts

问题 This is a mystery that I was trying to figure out when I was 15, but I failed. I still don't know the answer. Here's a naive and flawed solution (like some other failed attempts I've seen here on Stack Overflow): const numberToString = number => { let result = ''; let multiplier = Math.floor(Math.log10(number)); while (number > 0) { const currentDigit = Math.floor(number / 10 ** multiplier); if (multiplier === -1) result += '.'; result += `${currentDigit}`; number -= 10 ** multiplier *

问题 This is a mystery that I was trying to figure out when I was 15, but I failed. I still don't know the answer. Here's a naive and flawed solution (like some other failed attempts I've seen here on Stack Overflow): const numberToString = number => { let result = ''; let multiplier = Math.floor(Math.log10(number)); while (number > 0) { const currentDigit = Math.floor(number / 10 ** multiplier); if (multiplier === -1) result += '.'; result += `${currentDigit}`; number -= 10 ** multiplier *

问题 There are many algorithms or policies for finding a path with minimum or maximum costs. But, it is hard to find an approach that can find a path within (or below) a required cost (RC), i.e., such an RC is not a minimum or maximum one, and the actual cost should less than such an RC. I am looking for a feasible algorithm to find a path satisfying the two constraints: The cost of such a path should be lower than the required cost. The path from source to destination should contain as many hops