Title description
In the N * N grid, we placed some 1 * 1 * 1 cubes aligned with the x, y, and z axes. Each value v = grid[i][j] means v cubes are stacked on cell (i, j).
Now, we look at the projections of these cubes on the xy, yz, and zx planes.
The projection is like a shadow, mapping a three-dimensional body onto a two-dimensional plane. Here, when looking at the cube from the top, front and side, we will see the "shadow".
Returns the total area of all three projections.
Example
示例 1：
输入：[[2]]
输出：5
示例 2：
输入：[[1,2],[3,4]]
输出：17
解释：
这里有该形体在三个轴对齐平面上的三个投影(“阴影部分”)。
示例 3：
输入：[[1,0],[0,2]]
输出：8
示例 4：
输入：[[1,1,1],[1,0,1],[1,1,1]]
输出：14
示例 5：
输入：[[2,2,2],[2,1,2],[2,2,2]]
输出：21
prompt
1 <= grid.length = grid[0].length <= 50
0 <= grid[i][j] <= 50
Problem solving ideas
This question is easy to handle if you understand the mathematical meaning of projection. The so-called projection is the largest area
that can be cast on a certain plane, so the projection on the xoy plane is the number of non-zero items. I used gridSize*(*gridColSize)-the number of zero items To calculate
the projection on the xoz plane is the sum of the maximum value of each sub-array.
The projection on the yoz plane is the
sum of the maximum value of each sub-array at the same position. Adding the area of the three planes can
be improved: use an array to store the maximum value, Avoiding two nested loops
Code
int projectionArea(int** grid, int gridSize, int* gridColSize){
int xyCount=gridSize*(*gridColSize);
int xzCount=0;
int yzCount=0;
for(int i=0;i<gridSize;i++){
int xMax=grid[i][0];
for(int j=0;j<*gridColSize;j++){
if(grid[i][j]==0)xyCount--;
if(grid[i][j]>xMax)xMax=grid[i][j];
}xzCount+=xMax;
}
for(int p=0;p<*gridColSize;p++){
int yMax=grid[0][p];
for(int q=0;q<gridSize;q++){
if(grid[q][p]>yMax)yMax=grid[q][p];
} yzCount+=yMax;
}return xyCount+xzCount+yzCount;
}