883. Projection Area of 3D Shapes-三维形体投影面积

You are given an n x n grid where we place some 1 x 1 x 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 the cell (i, j).

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

Example 2:
Input: grid = [[2]]
Output: 5

Example 3:
Input: grid = [[1,0],[0,2]]
Output: 8

Example 4:
Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 14

Example 5:
Input: grid = [[2,2,2],[2,1,2],[2,2,2]]
Output: 21

Constraints:

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

Solution in python:

class Solution:
    def projectionArea(self, grid: List[List[int]]) -> int:
        top = 0
        front = 0
        side = 0
        for i in range(len(grid)):
            front += max(grid[i])
            for j in range(len(grid[i])):
                if grid[i][j] != 0:
                    top += 1

        for j in range(len(grid[0])):
            max_side = grid[0][j]
            for i in range(len(grid)):
                if grid[i][j] > max_side:
                    max_side = grid[i][j]
            side += max_side
        return top+front+side