puzzle

问题 What is the fastest algorithm for circle shifting array for M positions? For example, [3 4 5 2 3 1 4] shift M = 2 positions should be [1 4 3 4 5 2 3] . Thanks a lot. 回答1: If you want O(n) time and no extra memory usage (since array was specified), use the algorithm from Jon Bentley's book, "Programming Pearls 2nd Edition". It swaps all the elements twice. Not as fast as using linked lists but uses less memory and is conceptually simple. shiftArray( theArray, M ): size = len( theArray ) assert

问题 What is a non recursive algorithm for deciding whether a passed in amount can be built additively from a set of numbers. In my case I'm determining whether a certain currency amount (such as $40) can be met by adding up some combination of a set of bills (such as $5, $10 and $20 bills). That is a simple example, but the algorithm needs to work for any currency set (some currencies use funky bill amounts and some bills may not be available at a given time). So $50 can be met with a set of ($20

问题 I am trying to find out a block of same colored region starting from the top left corner in a 2D matrix. For example: I have the following Matrix: 1 1 1 2 2 3 1 1 2 3 4 5 1 1 1 1 3 4 1 4 3 2 1 5 2 3 4 5 1 2 say, The initial top left corner is 1 and I want to find out the adjacent area containing 1's (I will only consider starting from the top left corner). In the above matrix the digits 1,2,3,4,5 represent different colors. I have tried using the following code segment for finding out the

问题 I am solving the following problem from Codility: A small frog wants to get to the other side of a river. The frog is currently located at position 0, and wants to get to position X. Leaves fall from a tree onto the surface of the river. You are given a non-empty zero-indexed array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in minutes. The goal is to find the earliest time when the frog can jump to the

问题 Ok, so I have a 3 x 3 jig saw puzzle game that I am writing and I am stuck on the solution method. public Piece[][] solve(int r, int c) { if (isSolved()) return board; board[r][c] = null; for (Piece p : pieces) { if (tryInsert(p, r, c)) { pieces.remove(p); break; } } if (getPieceAt(r, c) != null) return solve(nextLoc(r, c).x, nextLoc(r, c).y); else { pieces.add(getPieceAt(prevLoc(r, c).x, prevLoc(r, c).y)); return solve(prevLoc(r, c).x, prevLoc(r, c).y); } } I know I haven't provided much

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 9 years ago . Is there any way that I can optimize this code as to not run out of memory? import java.util.HashMap; import java.util.Map; import java.util.PriorityQueue; import java.util.Random; import java.util.Stack; public

问题 I only learn Java Swing 1 week so I tried complete some exercises. This is my code. I set 9 icon to 9 button but it doesn't show on button. package mypack; import java.awt.Color; import javax.swing.*; public class PuzzleGame extends JFrame{ static JButton bt1,bt2,bt3,bt4,bt5,bt6,bt7,bt8,bt9,btNew,btExit; static JLabel move, moveNum, time, timeNum; public PuzzleGame(){ createMyGUI(); } public static void createMyGUI(){ JFrame jf = new JFrame("Game Puzzle Java"); JPanel jpl = new JPanel(); Icon

问题 Maybe I've had too much coffee, maybe I've been working too long, regardless, I'm at a loss as to what this method does, or rather, why and how it does it, could anyone shed some light upon me? What is the nextColor? public Color nextColor() { int max = 0, min = 1000000000, cr = 0, cg = 0, cb = 0; for (int r = 0; r < 256; r += 4) { for (int g = 0; g < 256; g += 4) { for (int b = 0; b < 256; b += 4) { if (r + g + b < 256 || r + g + b > 512) { continue; } min = 1000000000; for (Color c :

问题 I want to find the next survivor after a given position and number of people. (define renumber (lambda (position n) (if (< position 3) (+ position (- n 3)) (- position 3)))) (define survives? (lambda (position n) (if (< n 3) #t (if (= position 3) #f (survives? (renumber position n) (- n 1)))))) (define first-survivor-after (lambda (position n) (cond ((and (<= n 3)(<= position 3)) null) ((or (>= n 3)(>= position 3))(survives? position n) (if = #f survives?) (survives? (+ 1 position) n)

问题 I'm developing a game named woodntDie. This game consists in 9 pieces which should look like this: The aim is to arrange all those pieces and make it look like a dice: This is the four final possible solutions with all those 9 pieces. I'm trying to implement some restrictions but those arent enough: sum([V5, V6, V11, V12, V17, V18, V23, V24, V29, V30, V35, V36, V41, V42, V47, V48, V53, V54], #=, 7), The sum of the opposite faces should always be 7. In this restriction I'm telling that the sum