LeetCode-Projection Area of 3D Shapes

最新推荐文章于 2020-02-24 22:19:13 发布

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最新推荐文章于 2020-02-24 22:19:13 发布

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分类专栏： LeetCode 文章标签： LeetCode Java

本文链接：https://blog.youkuaiyun.com/qq_24133491/article/details/82558616

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本文介绍了一种计算三维立方体在三个不同平面上投影总面积的方法。通过分析给定的N*N网格中每个位置的立方体数量，文章提供了一个高效的算法来计算俯视图、主视图和左视图的立方体总数。

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Description：
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 2Input: [[1,2],[3,4]]
Output: 17xplanation:**
Here are the three projections (“shadows”) of the shape made with each axis-aligned plane.
这里写图片描述

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

题意：给定一个N*N的网格，每个位置的数字表示此位置有多少个1*1*1的立方体；现在要求计算这N*N的网格上所有立方体的俯视图、主视图和左视图的所有立方体个数；

解法：计算俯视图可以通过判断此位置上的立方体是否不为零；计算主视图可以累加每列的最大值；计算左视图可以累加每行的最大值；

class Solution {
    public int projectionArea(int[][] grid) {
        int top = 0;
        int front = 0;
        int side = 0;

        //get top and side
        for (int i = 0; i < grid.length; i++) {
            int max = 0;
            for (int j = 0; j < grid[0].length; j++) {
                top = grid[i][j] == 0 ? top : top + 1;
                max = grid[i][j] > max ? grid[i][j] : max;
            }
            side += max;
        }
        //get front
        for (int i = 0; i < grid[0].length; i++) {
            int max = 0;
            for (int j = 0; j < grid.length; j++) {
                max = grid[j][i] > max ? grid[j][i] : max;
            }
            front += max;
        }

        return top + front + side;
    }
}