[leetcode] 883. Projection Area of 3D Shapes

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

代码：

class Solution {
    public int projectionArea(int[][] grid) {
        int top = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[i].length; j++){
                top += grid[i][j] > 0 ? 1 : 0;
            }
        }
    
        int[] col = new int[grid.length];
        for(int i = 0; i < grid[0].length; i++){
            for(int j = 0; j < grid.length; j++){
                if(col[j] < grid[j][i]) col[j] = grid[j][i];
            }
        }
        for(int i = 0; i < col.length; i++){
            top += col[i];
        }
        
        
        int row[] = grid[0].clone();
        for(int i = 1; i < grid.length; i++){
            for(int j = 0; j < grid[i].length; j++){
                if(row[j] < grid[i][j]) row[j] = grid[i][j];
            }
        }
        for(int i = 0; i < row.length; i++){
            top += row[i];
        }
        return top;
    }
}