Title description
On the N * N grid, we place some 1 * 1 * 1 cubes.
Each value v = grid[i][j] means v cubes are stacked on the corresponding cell (i, j).
Please return the surface area of the final form.
Example
示例 1：
输入：[[2]]
输出：10
示例 2：
输入：[[1,2],[3,4]]
输出：34
示例 3：
输入：[[1,0],[0,2]]
输出：16
示例 4：
输入：[[1,1,1],[1,0,1],[1,1,1]]
输出：32
示例 5：
输入：[[2,2,2],[2,1,2],[2,2,2]]
输出：46
prompt
1 <= N <= 50
0 <= grid[i][j] <= 50
Problem solving ideas
Calculate the total area and subtract the hidden surface area. Top
and bottom coverage, left and right coverage, front and back coverage each -2
Code
int surfaceArea(int** grid, int gridSize, int* gridColSize){
int s=0;
int k;
for(int i=0;i<gridSize;i++){
for(int j=0;j<*gridColSize;j++){
for(k=0; k<grid[i][j]; k++){
s+=6;//一个立方体表面积为6
if(k>0){
s -= 2;
}
// 前面相邻
if(i>0 && grid[i-1][j]>k){
s -= 2;
}
// 左面相邻
if(j>0 && grid[i][j-1]>k){
s -= 2;
}
}
}
}
return s;
}