leetcode讲解--892. Surface Area of 3D Shapes

最新推荐文章于 2025-09-12 17:31:31 发布

weixin_33739541

最新推荐文章于 2025-09-12 17:31:31 发布

阅读量71

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法 java

原文链接：https://segmentfault.com/a/1190000017814941

本文详细解析了一道关于计算三维网格上放置立方体后形成的形状总表面积的问题，通过分类讨论的方法，给出了Java代码实现，适合对算法和数据结构有兴趣的读者。

题目

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

题目地址

讲解

这道题感觉还挺难的，不应该放在简单题里面。主要是考察分类讨论的能力，corner case非常之多，对分类讨论的能力要求很高。调试好久才过。

java代码

class Solution {
    public int surfaceArea(int[][] grid) {
        int result = 0;
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid.length;j++){
                if(i==0){
                    if(j==0){
                        if(grid[i][j]<=1){
                            result += 6*grid[i][j];
                        }else{
                            result += 6+4*(grid[i][j]-1);
                        }
                    }else{
                        if(grid[i][j]<=1){
                            result += grid[i][j]*6-2*(Math.min(grid[i][j],grid[i][j-1]));
                        }else{
                            result += 6+4*(grid[i][j]-1)-2*(Math.min(grid[i][j],grid[i][j-1]));
                        }
                    }
                }else{
                    if(j==0){
                        if(grid[i][j]<=1){
                            result += grid[i][j]*6-2*(Math.min(grid[i][j],grid[i-1][j]));
                        }else{
                            result += 6+4*(grid[i][j]-1)-2*(Math.min(grid[i][j],grid[i-1][j]));
                        }
                    }else{
                        if(grid[i][j]<=1){
                            result += grid[i][j]*6-2*(Math.min(grid[i][j],grid[i-1][j]))-2*(Math.min(grid[i][j],grid[i][j-1]));
                        }else{
                            result += 6+4*(grid[i][j]-1)-2*(Math.min(grid[i][j],grid[i-1][j]))-2*(Math.min(grid[i][j],grid[i][j-1]));
                        }
                    }
                }
            }
        }
        return result;
    }
}