LeetCode-Projection Area of 3D Shapes

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Description：
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 2Input: [[1,2],[3,4]]
Output: 17xplanation:**
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

题意：给定一个N*N的网格，每个位置的数字表示此位置有多少个1*1*1的立方体；现在要求计算这N*N的网格上所有立方体的俯视图、主视图和左视图的所有立方体个数；

解法：计算俯视图可以通过判断此位置上的立方体是否不为零；计算主视图可以累加每列的最大值；计算左视图可以累加每行的最大值；

class Solution {
    public int projectionArea(int[][] grid) {
        int top = 0;
        int front = 0;
        int side = 0;

        //get top and side
        for (int i = 0; i < grid.length; i++) {
            int max = 0;
            for (int j = 0; j < grid[0].length; j++) {
                top = grid[i][j] == 0 ? top : top + 1;
                max = grid[i][j] > max ? grid[i][j] : max;
            }
            side += max;
        }
        //get front
        for (int i = 0; i < grid[0].length; i++) {
            int max = 0;
            for (int j = 0; j < grid.length; j++) {
                max = grid[j][i] > max ? grid[j][i] : max;
            }
            front += max;
        }

        return top + front + side;
    }
}