You are given an?n x n?grid?where you have placed some?1 x 1 x 1?cubes. Each value?v = grid[i][j]?represents a tower of?v?cubes placed on top of cell?(i, j).

After placing these cubes, you have decided to glue any directly adjacent cubes to each other, forming several irregular 3D shapes.

Return?the total surface area of the resulting shapes.

Note:?The bottom face of each shape counts toward its surface area.

?

Example 1:

Input: grid = [[1,2],[3,4]]Output: 34

Example 2:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]Output: 32

Example 3:

Input: grid = [[2,2,2],[2,1,2],[2,2,2]]Output: 46

?

Constraints:

• n == grid.length == grid[i].length

• 1 <= n <= 50

• 0 <= grid[i][j] <= 50

Accepted

27,339

Submissions

44,530

橫排重復(fù)(i)，縱排重復(fù)(j)，上下重復(fù)(i,j)>1時候的面算出來，然后正方體的總數(shù)*6-重復(fù)的面就能得出；

Success

Details?

Runtime:?3 ms, faster than?83.66%?of?Java?online submissions for?Surface Area of 3D Shapes.

Memory Usage:?45 MB, less than?7.78%?of?Java?online submissions for?Surface Area of 3D Shapes.