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108 lines
1.8 KiB
Markdown
108 lines
1.8 KiB
Markdown
# [892. Surface Area of 3D Shapes](https://leetcode.com/problems/surface-area-of-3d-shapes/)
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## 题目
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On a `N * N` grid, we place some `1 * 1 * 1` cubes.
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Each value `v = grid[i][j]` represents a tower of `v` cubes placed on top of grid cell `(i, j)`.
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Return the total surface area of the resulting shapes.
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**Example 1**:
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```
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Input: [[2]]
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Output: 10
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```
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**Example 2**:
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```
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Input: [[1,2],[3,4]]
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Output: 34
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```
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**Example 3**:
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```
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Input: [[1,0],[0,2]]
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Output: 16
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```
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**Example 4**:
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```
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Input: [[1,1,1],[1,0,1],[1,1,1]]
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Output: 32
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```
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**Example 5**:
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```
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Input: [[2,2,2],[2,1,2],[2,2,2]]
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Output: 46
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```
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**Note**:
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- `1 <= N <= 50`
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- `0 <= grid[i][j] <= 50`
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## 题目大意
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在 N * N 的网格上，我们放置一些 1 * 1 * 1 的立方体。每个值 v = grid[i][j] 表示 v 个正方体叠放在对应单元格 (i, j) 上。请你返回最终形体的表面积。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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func surfaceArea(grid [][]int) int {
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area := 0
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for i := 0; i < len(grid); i++ {
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for j := 0; j < len(grid[0]); j++ {
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if grid[i][j] == 0 {
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continue
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}
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area += grid[i][j]*4 + 2
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// up
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if i > 0 {
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m := min(grid[i][j], grid[i-1][j])
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area -= m
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}
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// down
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if i < len(grid)-1 {
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m := min(grid[i][j], grid[i+1][j])
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area -= m
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}
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// left
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if j > 0 {
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m := min(grid[i][j], grid[i][j-1])
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area -= m
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}
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// right
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if j < len(grid[i])-1 {
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m := min(grid[i][j], grid[i][j+1])
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area -= m
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}
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}
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}
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return area
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}
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func min(a, b int) int {
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if a > b {
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return b
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}
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return a
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}
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``` |