【LeetCode】883. Projection Area of 3D Shapes
题目地址：https://leetcode.com/problems/projection-area-of-3d-shapes/
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:
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 <= 500 <= grid[i][j] <= 50
题意是一个N*N的方格，在每个格子上放一定数量的方块，输入就是每个格子上放的方块数量。
求它三视图所有的阴影部分的面积。
解题思路：
俯视图就是求每个格子上方块数不为0的数量。
主视图和左视图均是求行或列上最高的方块数。
class Solution {
public:
int projectionArea(vector<vector<int>>& grid) {
int top=0,front=0 ,side=0,N=grid[0].size();
for(int i=0;i<N;i++){
int max1=0,max2=0;
for(int j=0;j<N;j++){
if(grid[i][j]>0)top++;
max1=max(grid[i][j],max1);
max2=max(grid[j][i],max2);
}
front+=max1;
side+=max2;
}
return front+side+top;
}
};