LeetCode-Surface Area of 3D Shapes

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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 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

题意：给定一个二维数组，每个元素表示当前位置立方体的个数（每个立方体为111），要求计算二维数组中所有立方体在三维空间中的表面积；

解法：对于每个位置的立方体来说，如果高度为h，那么这个位置的表面积为S₁=h * 6 - 2 * (h - 1)；如果，这个立方体与周围的立方体是相邻的，我们还要扣除与之相邻的那部分表面积；此时，S₁ -= 前、后、左、右与之相邻的表面积；最后的表面积Area = S₁ + S₂ + …+ S_n；

Java

class Solution {
    public int surfaceArea(int[][] grid) {
        int area = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == 0) continue;
                area += 6 * grid[i][j] - 2 * (grid[i][j] - 1);
                area -= reduceAround(grid, i, j);
            }
        }
        return area;
    }
    
    private int reduceAround(int[][] grid, int row, int col) {
        int reduce = 0;
        if (row - 1 >= 0) reduce += Math.min(grid[row][col], grid[row - 1][col]);
        if (row + 1 < grid.length) reduce += Math.min(grid[row][col], grid[row + 1][col]);
        if (col - 1 >= 0) reduce += Math.min(grid[row][col], grid[row][col - 1]);
        if (col + 1 < grid[row].length) reduce += Math.min(grid[row][col], grid[row][col + 1]);
        return reduce;
    }
}