Solve a Bo asked questions, quite simple, write down to meet their own small sense of accomplishment

First issue: http://q.cnblogs.com/q/77610/

public static string[] Stops = new string[] {"AB", "BC", "CD", "DC", "DE", "AD", "CE", "EB", "AE"}; Algorithm: Enumerates a collection of paths between any 2 letters contained in the array. For example: A-c  has ABC (AB-BC), ADC (AD-DC)

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I have only learned simple data structure, so the simple problem to take out Ah, forgive me ~

According to the problem, can be abstracted as, give you a set of points and points between the connection, and then give you a starting point and the end point, you have all the matching path printed out is finished.

Very simple, with a map ah, but the premise is that the input data can not form a ring, or the program will loop up the ~

Post code:

1  Packagecnblogs;2 3 ImportJava.util.Stack;4 5 /**6 * Created by Kischn on 2015/11/21.7  */8  Public classTest {9 Ten     /** One * Map with direction A * The data must be constructed to ensure that there is no ring -      */ -      Public Static Boolean[] m =New Boolean[26] [26]; the  -     /** - * Print your way through -      * @param the +      */ -      Public Static voidPrintway (stack<integer>) { +          for(Integer node:way) { ASystem.out.print ((Char) (node + ' A ')); at         } - System.out.println (); -     } -  -     /** - * Find the end point and record the path to go in      * @paramStack -      * @param the to      * @paramEnd +      */ -      Public Static voidFind (stack<integer> Stack, stack<integer>,intend) { the         if(Stack.isempty ())return; *Integer tmp =Stack.pop (); $ Way.add (TMP);Panax Notoginseng         if(TMP = = end) {//End node found - Printway (to); the}Else { +              for(inti = 0; I < 26; i++) { A                 if(M[tmp][i]) { the Stack.push (i); + find (Stack, it, end); -                 } $             } $         } - Way.pop (); -     } the  -      Public Static voidMain (string[] args) {WuyiString[] Stops =NewString[] {"AB", "BC", "CD", "DC", "DE", "AD", "CE", "EB", "AE" }; the         CharStart = ' A '; -         Charend = ' C '; Wu  -         Charb; About          for(inti = 0; i < stops.length; i++){ $A = Stops[i].charat (0); -b = Stops[i].charat (1); -M[a-' A '][b-' a '] =true; -         } AStack<integer>NewStack<integer>(); +stack<integer> stack =NewStack<integer>(); theStack.push (Start-' A '); -Find (Stack, the, end-' A '); $     } the  the}

Solve a Bo asked questions, quite simple, write down to meet their own small sense of accomplishment