On a N * N
grid, we place some 1 * 1 * 1
cubes that are axis-aligned with the x, y, and z axes.
Each value v = grid[i][j]
represents a tower of v
cubes placed on top of grid cell (i, j)
.
Now we view the projection of these cubes onto the xy, yz, and zx planes.
A projection is like a shadow, that maps our 3 dimensional figure to a 2 dimensional plane.
Here, we are viewing the "shadow" when looking at the cubes from the top, the front, and the side.
Return the total area of all three projections.
Example 1:
Input: [[2]] Output: 5
Example 2:
Input: [[1,2],[3,4]] Output: 17 Explanation: Here are the three projections ("shadows") of the shape made with each axis-aligned plane.
Example 3:
Input: [[1,0],[0,2]] Output: 8
Example 4:
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Input: [[1,1,1],[1,0,1],[1,1,1]] Output: 14
Example 5:
Input: [[2,2,2],[2,1,2],[2,2,2]] Output: 21
Note:
1 <= grid.length = grid[0].length <= 50
0 <= grid[i][j] <= 50
代码:
class Solution {
public int projectionArea(int[][] grid) {
int top = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[i].length; j++){
top += grid[i][j] > 0 ? 1 : 0;
}
}
int[] col = new int[grid.length];
for(int i = 0; i < grid[0].length; i++){
for(int j = 0; j < grid.length; j++){
if(col[j] < grid[j][i]) col[j] = grid[j][i];
}
}
for(int i = 0; i < col.length; i++){
top += col[i];
}
int row[] = grid[0].clone();
for(int i = 1; i < grid.length; i++){
for(int j = 0; j < grid[i].length; j++){
if(row[j] < grid[i][j]) row[j] = grid[i][j];
}
}
for(int i = 0; i < row.length; i++){
top += row[i];
}
return top;
}
}