883. Projection Area of 3D Shapes*
https://leetcode.com/problems/projection-area-of-3d-shapes/
题目描述
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 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
C++ 实现 1
那第一个例子来说明:
class Solution {
public:
int projectionArea(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
int sum = 0;
// 向前看
for (auto &v : grid)
sum += *std::max_element(v.begin(), v.end());
// 侧向看
for (int j = 0; j < n; ++ j) {
auto val = 0;
for (int i = 0; i < m; ++ i)
val = std::max(val, grid[i][j]);
sum += val;
}
// 向下看
for (int i = 0; i < m; ++ i)
for (int j = 0; j < n; ++ j)
if (grid[i][j] > 0) sum += 1;
return sum;
}
};