题目:
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: 17Example 3:
Input: [[1,0],[0,2]]
Output: 8Example 4:
Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 14Example 5:
Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 21
解释:
用一个函数判断是否是self-dividing。
python代码:
class Solution(object):
def projectionArea(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
aera_1,aera_2,aera_3=0,0,0
N=len(grid)
for i in range(N):
tmp=[]
aera_2+=max(grid[i])
for j in range(N):
if grid[i][j]:
aera_1+=1
tmp.append(grid[j][i])
aera_3+=max(tmp)
return aera_1+aera_2+aera_3
c++代码:
class Solution {
public:
int projectionArea(vector<vector<int>>& grid) {
int area_1=0,area_2=0,area_3=0;
int N=grid.size();
for(int i=0;i<N;i++)
{
area_1+=*(max_element(grid[i].begin(),grid[i].end()));
vector<int>tmp;
for (int j=0;j<N;j++)
{
if (grid[i][j])
area_2+=1;
tmp.push_back(grid[j][i]);
}
area_3+=*(max_element(tmp.begin(),tmp.end()));
}
return area_1+area_2+area_3;
}
};
总结:
学会使用 max_element()函数,注意 max_element()函数返回的是指针,要求最大值时需要*,cpp的值一定要初始化,不然是随机值而不是0,这个一定要注意哟~