0892. Surface Area of 3 D Shapes

# 892. Surface Area of 3D Shapes#

## 题目 #

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`

## 解题思路 #

• 给定一个网格数组，数组里面装的是立方体叠放在所在的单元格，求最终这些叠放的立方体的表面积。
• 简单题。按照题目意思，找到叠放时，重叠的面，然后用总表面积减去这些重叠的面积即为最终答案。

## 代码 #

``````
package leetcode

func surfaceArea(grid [][]int) int {
area := 0
for i := 0; i < len(grid); i++ {
for j := 0; j < len(grid[0]); j++ {
if grid[i][j] == 0 {
continue
}
area += grid[i][j]*4 + 2
// up
if i > 0 {
m := min(grid[i][j], grid[i-1][j])
area -= m
}
// down
if i < len(grid)-1 {
m := min(grid[i][j], grid[i+1][j])
area -= m
}
// left
if j > 0 {
m := min(grid[i][j], grid[i][j-1])
area -= m
}
// right
if j < len(grid[i])-1 {
m := min(grid[i][j], grid[i][j+1])
area -= m
}
}
}
return area
}

func min(a, b int) int {
if a > b {
return b
}
return a
}

``````

Apr 8, 2023