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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;
}
}

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