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: 5Example 2:Input: [[1,2],[3,4]]Output: 17Explanation:Here are the three projections (“shadows”) of the shape made with each axis-aligned plane.Example 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: 21Note:1 <= grid.length = grid[0].length <= 500 <= 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; }}