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

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;
    }
}