LeetCode 892. Surface Area of 3D Shapes

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本文介绍了一个算法问题，即在一个N*N的网格上放置一些1*1*1的立方体，每个网格单元的值代表该位置立方体的数量。文章通过示例详细解释了如何计算最终形状的总表面积，并提供了一段Python代码实现。

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

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It's easy to be affected by the first question:

class Solution:
    def surfaceArea(self, grid: List[List[int]]) -> int:
        N,res = len(grid),0
        rmax,cmax = [0 for i in range(N)],[0 for i in range(N)]
        for i in range(N):
            for j in range(N):
                cur = grid[i][j]
                if (cur > 0):
                    res += cur*4+2
                if (i > 0):
                    res -= min(cur,grid[i-1][j])*2
                if (j > 0):
                    res -= min(cur,grid[i][j-1])*2  
        return res