892 Surface Area of 3D Shapes 三维形体的表面积
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.
Example:
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
题目描述:
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。
示例 :
示例 1：
输入：[[2]]
输出：10
示例 2：
输入：[[1,2],[3,4]]
输出：34
示例 3：
输入：[[1,0],[0,2]]
输出：16
示例 4：
输入：[[1,1,1],[1,0,1],[1,1,1]]
输出：32
示例 5：
输入：[[2,2,2],[2,1,2],[2,2,2]]
输出：46
提示：
1 <= N <= 50
0 <= grid[i][j] <= 50
思路:
单独看每一个立方体表面积是 4 * grid[i][j] + 2
再减去两个相邻的立方体重叠的部分面积
时间复杂度O(n ^ 2), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int surfaceArea(vector>& grid)
{
int result = 0;
for (int i = 0; i < grid.size(); i++) for (int j = 0; j < grid[0].size(); j++)
{
if (grid[i][j]) result += (grid[i][j] << 2) + 2;
if (i) result -= (min(grid[i - 1][j], grid[i][j]) << 1);
if (j) result -= (min(grid[i][j - 1], grid[i][j]) << 1);
}
return result;
}
};
Java:
class Solution {
public int surfaceArea(int[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] != 0) result += (grid[i][j] << 2) + 2;
if (i > 0) result -= (Math.min(grid[i - 1][j], grid[i][j]) << 1);
if (j > 0) result -= (Math.min(grid[i][j - 1], grid[i][j]) << 1);
}
}
return result;
}
}
Python:
class Solution:
def surfaceArea(self, grid: List[List[int]]) -> int:
return sum(((item << 2) + 2 for row in grid for item in row if item)) - sum(((min(row[i], row[i + 1]) << 1) for row in grid for i in range(len(row) - 1))) - sum(((min(column[j], column[j + 1]) << 1) for column in zip(*grid) for j in range(len(column) - 1)))