有效的方法来找到有多少个相邻元素与一个特定的值有一个特定的元素在一个数组中

Question

如果我有一个名为myArray的2D int数组，例如3x3大小，其元素只能有1或0值，那么计算特定元素周围有多少个1的有效方法是什么？ E.g：

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

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

一个neighbourCount这是我的当前蛮力代码。 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

怎么样：

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; 
}