883. Projection Area of 3D Shapes

#### QUESTION:

On a `N * N` grid, we place some `1 * 1 * 1 `cubes that are axis-aligned with the x, y, and z axes.

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

Now we view the projection of these cubes onto the xy, yz, and zx planes.

A projection is like a shadow, that maps our 3 dimensional figure to a 2 dimensional plane.

Here, we are viewing the “shadow” when looking at the cubes from the top, the front, and the side.

Return the total area of all three projections.

Example 1:

``````Input: [[2]]
Output: 5
``````

Example 2:

``````Input: [[1,2],[3,4]]
Output: 17
Explanation:
Here are the three projections ("shadows") of the shape made with each axis-aligned plane.
``````

Example 3:

``````Input: [[1,0],[0,2]]
Output: 8
``````

Example 4:

``````Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 14
``````

Example 5:

``````Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 21
``````

Note:

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

#### EXPLANATION:

x方向的投影，其实就是所有的yz这个平面上的最大值。

y方向的投影，就是所有xz这个平面上的值。

z方向的投影，就是xy平面上的值。

#### SOLUTION:

``````class Solution {
public int projectionArea(int[][] grid) {
int sumX = 0;
int sumY = 0;
int sumZ = 0;

for(int i =0;i<grid.length;i++){
int tmpY = 0;
int tmpX = 0;
int tmpZ = 0;
for(int j=0;j<grid[i].length;j++){
tmpY = Math.max(tmpY,grid[i][j]);
tmpX = Math.max(tmpX,grid[j][i]);
if(grid[i][j]!=0) tmpZ++;
}
sumX+=tmpX;
sumY+=tmpY;
sumZ+=tmpZ;
}
return sumX+sumY+sumZ;
}
}
``````