How to Compute the Surface Area of 3D Shapes (Cubes Placed on Gr
- 时间:2020-09-18 17:01:02
- 分类:网络文摘
- 阅读:89 次
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 1:
Input: [[2]]
Output: 10Example 2:
Input: [[1,2],[3,4]]
Output: 34Example 3:
Input: [[1,0],[0,2]]
Output: 16Example 4:
Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 32Example 5:
Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 46Note:
1 <= N <= 50
0 <= grid[i][j] <= 50
When we place a single cube on the grid, the surface is 6 (4×1+2), when we place a 2×1 cubes on the grid, the surface is 10 – which is 4×2+2. That is, there is only 1 top and 1 bottom, but 4 times of the number cubes that stacked together – as the connected parts (vertically) are hidden.
Then, we can iterate each verticl stacked cubes, add the top and bottom, count the side surfaces by checking the four neighbours, and add the difference.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | class Solution { public: int surfaceArea(vector<vector<int>>& grid) { int dr[4] = {0, 1, 0, -1}; int dc[4] = {1, 0, -1, 0}; int N = grid.size(); int ans = 0; for (int r = 0; r < N; ++ r) { for (int c = 0; c < N; ++ c) { if (grid[r][c] > 0) { ans += 2; for (int k = 0; k < 4; ++ k) { int nr = r + dr[k]; int nc = c + dc[k]; int nv = 0; if ((0 <= nr) && (0 <= nc) && (nr < N) && (nc < N)) { nv = grid[nr][nc]; } ans += max(grid[r][c] - nv, 0); } } } } return ans; } }; |
class Solution { public: int surfaceArea(vector<vector<int>>& grid) { int dr[4] = {0, 1, 0, -1}; int dc[4] = {1, 0, -1, 0}; int N = grid.size(); int ans = 0; for (int r = 0; r < N; ++ r) { for (int c = 0; c < N; ++ c) { if (grid[r][c] > 0) { ans += 2; for (int k = 0; k < 4; ++ k) { int nr = r + dr[k]; int nc = c + dc[k]; int nv = 0; if ((0 <= nr) && (0 <= nc) && (nr < N) && (nc < N)) { nv = grid[nr][nc]; } ans += max(grid[r][c] - nv, 0); } } } } return ans; } };
Slightly differently, we can check the neighbours of north and west only and minus those connected surfaces.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | class Solution { public: int surfaceArea(vector<vector<int>>& grid) { int N = grid.size(); int ans = 0; for (int r = 0; r < N; ++ r) { for (int c = 0; c < N; ++ c) { if (grid[r][c] > 0) { ans += 4 * grid[r][c] + 2; if (r > 0) ans -= min(grid[r][c], grid[r - 1][c]) * 2; if (c > 0) ans -= min(grid[r][c], grid[r][c - 1]) * 2; } } } return ans; } }; |
class Solution { public: int surfaceArea(vector<vector<int>>& grid) { int N = grid.size(); int ans = 0; for (int r = 0; r < N; ++ r) { for (int c = 0; c < N; ++ c) { if (grid[r][c] > 0) { ans += 4 * grid[r][c] + 2; if (r > 0) ans -= min(grid[r][c], grid[r - 1][c]) * 2; if (c > 0) ans -= min(grid[r][c], grid[r][c - 1]) * 2; } } } return ans; } };
Both algorithms/approaches are based on the counting, which result in O(N^2) time and O(1) space requirement.
Similar post: How to Compute the Projection Area of 3D Shapes?
–EOF (The Ultimate Computing & Technology Blog) —
推荐阅读:Innovative Plugins You Need On Your Mobile Business Site Why Having a Phone Service is a Must for Businesses? How to Validate a Perfect Number (Integer)? Binary Prefix Divisible By 5 – Java/C++ Coding Exercise The DNS Lookup Tool in Java (InetAddress) How to Check if a Matrix is a Toeplitz Matrix? How to Uncommon Words from Two Sentences in Java? The Best Bootable USB Creation Tool for Windows, Linux and Mac How to Compute the Number Complement for Integers? 4 Reasons People Skim Your Blog And How To Change That
- 评论列表
-
- 添加评论