892. Surface Area of 3D Shapes

#### QUESTION:

On a `N * N` grid, we place some `1 * 1 * 1 `cubes.

Each value `v = grid[i][j]` represents a tower of `v` cubes placed on top of grid cell `(i, j)`.

Return the total surface area of the resulting shapes.

Example 1:

``````Input: [[2]]
Output: 10
``````

Example 2:

``````Input: [[1,2],[3,4]]
Output: 34
``````

Example 3:

``````Input: [[1,0],[0,2]]
Output: 16
``````

Example 4:

``````Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 32
``````

Example 5:

``````Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 46
``````

Note:

• `1 <= N <= 50`
• `0 <= grid[i][j] <= 50`

#### EXPLANATION:

1.首先计算获得当前的整个面积

2.减去相邻四面可能重合的面积

1.比相邻的高，那么只需要加上减去相邻高度

2.比相邻的矮，那么就不用加

#### SOLUTION:

``````    public static int surfaceArea(int[][] grid) {
int[] dr = new int[]{1,-1,0,0};
int[] dc = new int[]{0,0,-1,1};
int result = 0;
for (int i = 0;i<grid.length;i++){
for(int j = 0;j<grid[i].length;j++){
if(grid[i][j]!=0){
result+=2;
for(int k = 0;k<4;k++){
int v = 0;
int nr = i + dr[k];
int nc = j + dc[k];
if(nr>=0&&nr<grid.length&&nc>=0&&nc<grid.length)
v = grid[nr][nc];

result+=Math.max(grid[i][j]-v,0);
}
}
}
}
return result;
}
``````
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