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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;
}
}
