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：134 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: 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

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.

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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.

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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) —

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