892. The three-dimensional body surface area
N * N on the grid, we put some cube 1 * 1 * 1.
Each value v = grid [i] [j] denotes the v cube superposed on the corresponding cell (i, j).
Please return to the final body surface area.
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
prompt:
1 <= N <= 50
0 <= grid[i][j] <= 50
Thinking
First, to separate the cut bars for each position, each of the bins of the surface area of 4n + 2. And then subtracting the overlapping area between the bins, where only consider the top left, to avoid double counting. Repeat area of min (n1, n2) * 2.
My answer
func surfaceArea(grid [][]int) int {
sf:=0
for i:=0;i<len(grid);i++{
for j:=0;j<len(grid[0]);j++{
if grid[i][j]!=0{
sf+=grid[i][j]*4+2
if i>0{
sf-=min(grid[i][j],grid[i-1][j])*2
}
if j>0{
sf-=min(grid[i][j],grid[i][j-1])*2
}
}
}
}
return sf
}
func min(x,y int) int {
if x<y{
return x
}
return y
}
