883 Projection Area of 3D Shapes 三维形体投影面积
Description:
On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, and z axes.
Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j).
Now we view the projection of these cubes onto the xy, yz, and zx planes.
A projection is like a shadow, that maps our 3 dimensional figure to a 2 dimensional plane.
Here, we are viewing the "shadow" when looking at the cubes from the top, the front, and the side.
Return the total area of all three projections.
Example:
Example 1:
Input: [[2]]
Output: 5
Example 2:
Input: [[1,2],[3,4]]
Output: 17
Explanation:
Here are the three projections ("shadows") of the shape made with each axis-aligned plane.
Example 3:
Input: [[1,0],[0,2]]
Output: 8
Example 4:
Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 14
Example 5:
Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 21
Note:
1 <= grid.length = grid[0].length <= 50
0 <= grid[i][j] <= 50
题目描述:
在 N * N 的网格中，我们放置了一些与 x，y，z 三轴对齐的 1 * 1 * 1 立方体。
每个值 v = grid[i][j] 表示 v 个正方体叠放在单元格 (i, j) 上。
现在，我们查看这些立方体在 xy、yz 和 zx 平面上的投影。
投影就像影子，将三维形体映射到一个二维平面上。
在这里，从顶部、前面和侧面看立方体时，我们会看到“影子”。
返回所有三个投影的总面积。
示例 :
示例 1：
输入：[[2]]
输出：5
示例 2：
输入：[[1,2],[3,4]]
输出：17
解释：
这里有该形体在三个轴对齐平面上的三个投影(“阴影部分”)。
示例 3：
输入：[[1,0],[0,2]]
输出：8
示例 4：
输入：[[1,1,1],[1,0,1],[1,1,1]]
输出：14
示例 5：
输入：[[2,2,2],[2,1,2],[2,2,2]]
输出：21
提示：
1 <= grid.length = grid[0].length <= 50
0 <= grid[i][j] <= 50
思路:
x方向上的投影是每个分量的最大值, y方向上的投影是转置之后的每个分量的最大值, z方向上的投影是该位置有方块( > 0), 记为 1
时间复杂度O(n ^ 2), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int projectionArea(vector>& grid)
{
int result = 0;
for (int i = 0; i < grid.size(); i++)
{
int r = 0, c = 0;
for (int j = 0; j < grid[i].size(); j++)
{
if (grid[i][j]) result++;
r = max(r, grid[i][j]);
c = max(c, grid[j][i]);
}
result += r + c;
}
return result;
}
};
Java:
class Solution {
public int projectionArea(int[][] grid) {
int result = 0;
for (int i = 0; i < grid.length; i++) {
int r = 0, c = 0;
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] > 0) result++;
r = Math.max(r, grid[i][j]);
c = Math.max(c, grid[j][i]);
}
result += r + c;
}
return result;
}
}
Python:
class Solution:
def projectionArea(self, grid: List[List[int]]) -> int:
return sum((max(i) for i in grid)) + sum((1 for i in grid for j in i if j)) + sum((max(i) for i in zip(*grid)))
# 评论解答
# return sum([sum(map(max, grid)), sum(map(max, zip(*grid))), sum(v > 0 for row in grid for v in row)])