On a
N * N
grid, we place some1 * 1 * 1
cubes.Each value
v = grid[i][j]
represents a tower ofv
cubes placed on top of grid cell(i, j)
.Return the total surface area of the resulting shapes.
Example 1:
Input: [[2]] Output: 10Example 2:
Input: [[1,2],[3,4]] Output: 34Example 3:
Input: [[1,0],[0,2]] Output: 16Example 4:
Input: [[1,1,1],[1,0,1],[1,1,1]] Output: 32Example 5:
Input: [[2,2,2],[2,1,2],[2,2,2]] Output: 46
Note:
1 <= N <= 50
0 <= grid[i][j] <= 50
Approach #1: Math. [Java]
class Solution { public int surfaceArea(int[][] grid) { int N = grid.length; int ret = 0; for (int i = 0; i < N; ++i) { for (int j = 0; j < N; ++j) { if (grid[i][j] > 0) ret += grid[i][j] * 4 + 2; if (i > 0) ret -= Math.min(grid[i][j], grid[i-1][j]) * 2; if (j > 0) ret -= Math.min(grid[i][j], grid[i][j-1]) * 2; } } return ret; } }
Analysis:
For each tower, its surface area is 4 * v + 2. However, 2 adjacent tower will hide the area of connected part. The hidden part is min(v1, v2) and we need just minus this area * 2.
Reference: