有效的方法來找到有多少個相鄰元素與一個特定的值有一個特定的元素在一個數組中

Question

如果我有一個名爲myArray的2D int數組，例如3x3大小，其元素只能有1或0值，那麼計算特定元素周圍有多少個1的有效方法是什麼？ E.g：

[0][0][0] 
[0][1][0] 
[1][1][1] 

元件myArray的[0] [0]將具有1 neighbourCount而myArray的[0] [1]將具有3.

一個neighbourCount這是我的當前蠻力代碼。 myArray的= currentGeneration

public int neighbours(int x, int y) { //x and y are 0 index based coordinates, they are swapped inside to corespond with actual x and y coordinates 
     int neighbourCounter = 0; 
     if(x == 0 && y == 0) { 
      if(currentGeneration[y+1][x] == 1) { 
       neighbourCounter++; 
      } 
      if(currentGeneration[y][x+1] == 1) { 
       neighbourCounter++; 
      } 
      if(currentGeneration[y+1][x+1] == 1) { 
       neighbourCounter++; 
      } 
     } else if(x == 0 && y == currentGeneration.length - 1) { 
          if(currentGeneration[y-1][x] == 1) { 
            neighbourCounter++; 
          } 
          if(currentGeneration[y][x+1] == 1) { 
            neighbourCounter++; 
          } 
          if(currentGeneration[y-1][x+1] == 1) { 
            neighbourCounter++; 
          } 
     } else if(x == currentGeneration[0].length - 1 && y == currentGeneration.length - 1) { 
          if(currentGeneration[y-1][x] == 1) { 
            neighbourCounter++; 
          } 
          if(currentGeneration[y][x-1] == 1) { 
            neighbourCounter++; 
          } 
          if(currentGeneration[y-1][x-1] == 1) { 
            neighbourCounter++; 
          } 
     } else if(y == 0 && x == currentGeneration[0].length - 1) { 
          if(currentGeneration[y][x-1] == 1) { 
            neighbourCounter++; 
          } 
          if(currentGeneration[y+1][x] == 1) { 
            neighbourCounter++; 
          } 
          if(currentGeneration[y+1][x-1] == 1) { 
            neighbourCounter++; 
          } 
     } else if(y == 0) { 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y][x+i] == 1) { 
        neighbourCounter++; 
       } 
      } 
      if(currentGeneration[y+1][x] == 1) { 
       neighbourCounter++; 
      } 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y+1][x+i] == 1) { 
        neighbourCounter++; 
       } 
      } 
     } else if(x == 0) { 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y+i][x] == 1) { 
        neighbourCounter++; 
       } 
      } 
      if(currentGeneration[y][x+1] == 1) { 
       neighbourCounter++; 
      } 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y+i][x+1] == 1) { 
        neighbourCounter++; 
       } 
      } 
     } else if(y == currentGeneration.length - 1) { 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y][x+i] == 1) { 
        neighbourCounter++; 
       } 
      } 
      if(currentGeneration[y-1][x] == 1) { 
       neighbourCounter++; 
      } 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y-1][x+i] == 1) { 
        neighbourCounter++; 
       } 
      } 
     } else if(x == currentGeneration[0].length - 1) { 
      for(int i = -1; i <= 2; i+=2) { 
       if(currentGeneration[y+i][x] == 1) { 
        neighbourCounter++; 
       } 
      } 
      if(currentGeneration[y][x-1] == 1) { 
       neighbourCounter++; 
      } 
      for(int i = -1; i <= 2; i+=2) { 
       if(currentGeneration[y+i][x-1] == 1) { 
        neighbourCounter++; 
       } 
      } 
     } else { 
      for(int i = -1; i <= 1; i+=2) { 
       if(currentGeneration[y+i][x] == 1) { 
        neighbourCounter++; 
       } 
       if(currentGeneration[y][x+i] == 1) { 
        neighbourCounter++; 
       } 
       if(currentGeneration[y+i][x+i] == 1) { 
        neighbourCounter++; 
       } 
       if(currentGeneration[y+i][x-i] == 1) { 
        neighbourCounter++; 
       } 
      } 
     } 
     return neighbourCounter; 
    } 

Answer 1

怎麼樣：

public static int findNeighbors(int x, int y, int[][] a) { 
    int sum = 0; 
    for (int i = (y>0 ? y-1 : 0); i <= (y<a.length-1 ? y+1 : a.length-1); ++i) 
     for (int j = (x>0 ? x-1 : 0); j <= (x<a[0].length-1 ? x+1 : a[0].length-1); ++j) 
      sum += a[i][j]; 
    sum -= a[y][x]; 
    return sum; 
}