### 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: 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 <= 50
• 0 <= grid[i][j] <= 50

### 问题的本质：求每行每列的最大值

1,2
3,4


1,2  -> max = 2
3,4  -> max = 3


1,2
3,4
| |
3 4


#### 代码

public int projectionArea(int[][] grid) {
int x = 0, y = 0, z = 0;
int[] maxY = new int[grid[0].length];   //记录每一列的最大值
for (int i = 0; i < grid.length; i++) {
int maxX = 0;                       //记录每一行的最大值
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] != 0) {
maxX = Math.max(grid[i][j],maxX);
maxY[j] = Math.max(grid[i][j],maxY[j]);
z++;
}
}
x += maxX;
}
for (int i = 0; i < grid[0].length; i++) {
y += maxY[i];
}
return x + y + z;
}