leetcode讲解--883. Projection Area of 3D Shapes

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

题目地址

讲解

这题其实很简单，就是考察对数组的横纵操作。

Java代码

class Solution {
    public int projectionArea(int[][] grid) {
        return countRow(grid)+countColumn(grid)+countPlan(grid);
    }
    
    private int countRow(int[][] grid){
        int result = 0;
        for(int i=0;i<grid.length;i++){
            int max = grid[i][0];
            for(int j=0;j<grid[i].length;j++){
                if(max<grid[i][j]){
                    max = grid[i][j];
                }
            }
            result+=max;
        }
        return result;
    }
    
    private int countColumn(int[][] grid){
        int result = 0;
        for(int i=0;i<grid[0].length;i++){
            int max = grid[0][i];
            for(int j=0;j<grid.length;j++){
                if(max<grid[j][i]){
                    max=grid[j][i];
                }
            }
            result+=max;
        }
        return result;
    }
    
    private int countPlan(int[][] grid){
        int result = 0;
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[i].length;j++){
                if(grid[i][j]>0){
                    result++;
                }
            }
        }
        return result;
    }
}