883. Projection Area of 3D Shapes*

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;
    }
};