如何查找數組中所有連接的數字？

Question

嘿，所有，再回來。在地下城發電機上工作，我對自己的進步感到驚訝。然而，我現在仍然有一個離散的房間。我想知道是否有一種方法可以循環訪問數組，並查看是否所有「1」（地磚）都已連接，如果沒有，則如何連接它們。

謝謝！

編輯：數組是隨機填充房間和走廊;下面的代碼：

import java.util.Random; 
public class Level 
{ 
    Random random = new Random(); 
    int[][] A = new int[100][100]; 
    int minimum = 3; 
    int maximum = 7; 
    int xFeature = 0; 
    int yFeature = 0; 
private void feature() 
    { 
    int i = 0; 
    while(i>=0) 
    { 
     xFeature = random.nextInt(100-1) + 1; 
     yFeature = random.nextInt(100-1) + 1; 
     if(A[xFeature][yFeature]==1)//||A[xFeature++][yFeature]==1||A[xFeature][yFeature--]==1||A[xFeature][yFeature++]==1) 
     break; 
     i++; 
    } 
    } 


    private void room() 
    { 
    int safeFall = 0; 
    int xCoPLUS = minimum + (int)(Math.random()*minimum); 
    int yCoPLUS = minimum + (int)(Math.random()*minimum); 
    if(yCoPLUS >= xCoPLUS) 
    { 
     for(int across = xFeature; across < xFeature+xCoPLUS+2; across++) 
     { 
     for(int vert = yFeature; vert < yFeature+yCoPLUS+1; vert++) 
     { 
      if(A[vert][across] == 0) 
      safeFall++; 
      else 
      break; 
     } 
     } 
    } 
    if(yCoPLUS < xCoPLUS) 
    { 
     for(int across = xFeature; across < xFeature+xCoPLUS+1; across++) 
     { 
     for(int vert = yFeature; vert < yFeature+yCoPLUS+2; vert++) 
     { 
      if(A[vert][across] == 0) 
      safeFall++; 
      else 
      break; 
     } 
     } 
    } 
    if((safeFall== (xCoPLUS+1) * (yCoPLUS+2)) || ((safeFall== (xCoPLUS+2) * (yCoPLUS+1)))) 
    { 
     for(int across = xFeature; across < xFeature+xCoPLUS; across++) 
     { 
     for(int vert = yFeature; vert < yFeature+yCoPLUS; vert++) 
     { 
      A[vert][across] = 1; 
     } 
     } 
    } 
    } 
private void corridor() 
    { 

    int xCoONE = xFeature; 
    int yCoONE = yFeature; 
    int xCoTWO = random.nextInt(10)+10; 
    int yCoTWO = random.nextInt(10)+10; 
    while(xCoONE > xCoTWO) 
    { 
     A[xCoONE][yCoONE] = 1; 
     xCoONE--; 
    } 
    while(xCoONE < xCoTWO) 
    { 
     A[xCoONE][yCoONE] = 1; 
     xCoONE++; 
    } 
    while(yCoONE > yCoTWO) 
    { 
     A[xCoONE][yCoONE] = 1; 
     yCoONE--; 
    } 
    while(yCoONE < yCoTWO) 
    { 
     A[xCoONE][yCoONE] = 1; 
     yCoONE++; 
    } 
} 
public Level() 
    { 
    firstroom(); 
    for(int i = 0; i < 500; i++) 
    { 
     int x = random.nextInt(50); 
     feature(); 
     if(x > 1) 
     room(); 
     else 
     corridor(); 
    } 
    troubleShoot(); 
    } 

所以基本上，當我創建這個類的一個對象是一個100x100的陣列充滿走廊和房間的隨機數確定會發生什麼。 （好吧，其中有幾個）但是我的房間沒有重疊的故障安全（safeFall在room（））中，我被一個房間卡住了，這個房間不時被觸及。

Answer 1

的製品Maze Generation Algorithm討論了幾種方法來生成一個迷宮。它包含了Java示例的鏈接。