LeetCode: 892. Surface Area of 3D Shapes

题目描述

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

解题思路

分别对每个每个坐标的前后左右上下六个方向求表面积。

AC 代码

class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int surfaceArea = 0;

        // 正视图
        for(int i = 0; i < grid.size(); ++i)
        {
            for(int j = 0; j < grid[i].size(); ++j)
            {
                //上下表面
                if(grid[i][j] > 0) surfaceArea += 2;

                // 左表面
                if(i - 1 >= 0 && grid[i-1][j] < grid[i][j]) 
                {
                    surfaceArea += (grid[i][j]-grid[i-1][j]);
                }
                else if(i - 1 < 0) surfaceArea += grid[i][j];

                // 右表面
                if(i + 1 < grid.size() && grid[i+1][j] < grid[i][j]) 
                {
                    surfaceArea += (grid[i][j]-grid[i+1][j]);
                }
                else if(i + 1 >= grid.size()) surfaceArea += grid[i][j];

                // 前表面
                if(j - 1 >= 0 && grid[i][j-1] < grid[i][j]) 
                {
                    surfaceArea += (grid[i][j]-grid[i][j-1]);
                }
                else if(j - 1 < 0) surfaceArea += grid[i][j];

                // 后表面
                if(j + 1 < grid.size() && grid[i][j+1] < grid[i][j]) 
                {
                    surfaceArea += (grid[i][j]-grid[i][j+1]);
                }
                else if(j + 1 >= grid.size()) surfaceArea += grid[i][j];
            }
        }

        return surfaceArea;
    }
};