883. Projection Area of 3D Shape

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.

image.png

思路

这个题首先要看懂grid数组是什么意思，比如Ex2,[[1, 2], [3, 4]]，表示在(x,y)为坐标的（0，0）处放了一个方块，在（0，1）处放了两个，依此类推。 可能我这个放置方法和题目中的放置有些许不同，但是转换到三视图中，方块的个数是一样多的。再看看计算方法，x,y方向就是找最大值，把数组写成下边形式容易看出

[ [1, 2],
  [3, 4]]

x,y方向就是行列分别找最大值，z方向是看数组的值，不为零，就加一。由于没有必要区分x，y轴，所以程序中就不必加以区分。

class Solution {
public:
    int projectionArea(vector<vector<int>>& grid) {
        int sum = 0;
        for(int i = 0; i < grid.size(); i++)
        {
            int tmp1 = grid[i][0];
            int tmp2 = grid[0][i];
            for(int j = 0; j < grid.size(); j++)//每行每列找最大值
            {
                if(grid[i][j]>tmp1)
                    tmp1 = grid[i][j];
                if(grid[j][i]>tmp2)
                    tmp2 = grid[j][i];
                if(grid[i][j]!=0)//查看元素是否为零
                    sum++;
            }
             sum += (tmp1+tmp2);
        }
        
        return sum;
        
    }
};