Description
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.
题目链接：https://leetcode.com/problems/surface-area-of-3d-shapes/description/
Difficulty: easy
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
分析
- 对于每一个立方体，由六个面，在整个立方体集合中贡献最大表面积为6；
- 如果上面有立方体则减一，如果下面有立方体则减一，以此类推（包含上下左右前后六个面）；
- 因为N和grid[i][j]都不是很大，所以便利所以立方体便可以得到表面积
参考代码
def surfaceArea(self, grid):
row=len(grid)
s=0
for i in range(row):
for j in range(len(grid[i])):
for index in range(grid[i][j]):
current=6
if(i-1>=0):
if(grid[i-1][j]>=index+1):
current-=1
if(i+1<row):
if(grid[i+1][j]>=index+1):
current-=1
if(j-1>=0):
if(grid[i][j-1]>=index+1):
current-=1
if(j+1<len(grid[i])):
if(grid[i][j+1]>=index+1):
current-=1
if(index>0):
current-=1
if(index < grid[i][j]-1):
current-=1
s+=current
return s
90 / 90 test cases passed.
Runtime: 496 ms