README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0892.Surface%20Area%20of%203D%20Shapes/README_EN.md	Geometry Array Math Matrix

892. Surface Area of 3D Shapes

中文文档

Description

You are given an n x n grid where you have placed some 1 x 1 x 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of cell (i, j).

After placing these cubes, you have decided to glue any directly adjacent cubes to each other, forming several irregular 3D shapes.

Return the total surface area of the resulting shapes.

Note: The bottom face of each shape counts toward its surface area.

 

Example 1:

Input: grid = [[1,2],[3,4]]
Output: 34

Example 2:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 32

Example 3:

Input: grid = [[2,2,2],[2,1,2],[2,2,2]]
Output: 46

 

Constraints:

• n == grid.length == grid[i].length
• 1 <= n <= 50
• 0 <= grid[i][j] <= 50

Solutions

Solution 1

Python3

class Solution:
    def surfaceArea(self, grid: List[List[int]]) -> int:
        ans = 0
        for i, row in enumerate(grid):
            for j, v in enumerate(row):
                if v:
                    ans += 2 + v * 4
                    if i:
                        ans -= min(v, grid[i - 1][j]) * 2
                    if j:
                        ans -= min(v, grid[i][j - 1]) * 2
        return ans

Java

class Solution {
    public int surfaceArea(int[][] grid) {
        int n = grid.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] > 0) {
                    ans += 2 + grid[i][j] * 4;
                    if (i > 0) {
                        ans -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
                    }
                    if (j > 0) {
                        ans -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
                    }
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int n = grid.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j]) {
                    ans += 2 + grid[i][j] * 4;
                    if (i) ans -= min(grid[i][j], grid[i - 1][j]) * 2;
                    if (j) ans -= min(grid[i][j], grid[i][j - 1]) * 2;
                }
            }
        }
        return ans;
    }
};

Go

func surfaceArea(grid [][]int) int {
	ans := 0
	for i, row := range grid {
		for j, v := range row {
			if v > 0 {
				ans += 2 + v*4
				if i > 0 {
					ans -= min(v, grid[i-1][j]) * 2
				}
				if j > 0 {
					ans -= min(v, grid[i][j-1]) * 2
				}
			}
		}
	}
	return ans
}