661 Image Smoother 图片平滑器
Description:
Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
Example:
Example 1:
Input:
[[1,1,1],
[1,0,1],
[1,1,1]]
Output:
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
Explanation:
For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0
For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0
For the point (1,1): floor(8/9) = floor(0.88888889) = 0
Note:
The value in the given matrix is in the range of [0, 255].
The length and width of the given matrix are in the range of [1, 150].
题目描述:
包含整数的二维矩阵 M 表示一个图片的灰度。你需要设计一个平滑器来让每一个单元的灰度成为平均灰度 (向下舍入) ，平均灰度的计算是周围的8个单元和它本身的值求平均，如果周围的单元格不足八个，则尽可能多的利用它们。
示例 :
示例 1:
输入:
[[1,1,1],
[1,0,1],
[1,1,1]]
输出:
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
解释:
对于点 (0,0), (0,2), (2,0), (2,2): 平均(3/4) = 平均(0.75) = 0
对于点 (0,1), (1,0), (1,2), (2,1): 平均(5/6) = 平均(0.83333333) = 0
对于点 (1,1): 平均(8/9) = 平均(0.88888889) = 0
注意：
给定矩阵中的整数范围为 [0, 255]。
矩阵的长和宽的范围均为 [1, 150]。
思路:
- 判断 8个方向进行计算, 注意边界条件即可
- 可以用偏移数组简化计算
- 在原数组外面包裹一层 None(或者小数, 能与原数据区分即可), 可以减少代码量
时间复杂度O(mn), 空间复杂度O(1), m为矩阵宽度, n为矩阵高度
代码:
C++:
class Solution
{
public:
vector> imageSmoother(vector>& M)
{
vector> result = M;
int row = M.size(), col = M[0].size();
for (int i = 0; i < row; i++) for (int j = 0; j < col; j++)
{
int count = 1, sum = M[i][j];
if (i > 0)
{
++count;
sum += M[i - 1][j];
if (j > 0)
{
++count;
sum += M[i - 1][j - 1];
}
if (j < col - 1)
{
++count;
sum += M[i - 1][j + 1];
}
}
if (i < row - 1)
{
++count;
sum += M[i + 1][j];
if (j > 0)
{
++count;
sum += M[i + 1][j - 1];
}
if (j < col - 1)
{
++count;
sum += M[i + 1][j + 1];
}
}
if (j > 0)
{
++count;
sum += M[i][j - 1];
}
if (j < col - 1)
{
++count;
sum += M[i][j + 1];
}
result[i][j] = sum / count;
}
return result;
}
};
Java:
class Solution {
public int[][] imageSmoother(int[][] M) {
int row = M.length, col = M[0].length, result[][] = new int[row][col];
for (int i = 0; i < row; i++) for (int j = 0; j < col; j++) result[i][j] = calculate(M, i, j, row, col);
return result;
}
private final int x[] = {-1, 1, 0, 0, -1, -1, 1, 1}, y[] = {0, 0, -1, 1, -1, 1, -1, 1};
private int calculate(int[][] M, int i, int j, int row, int col) {
int count = 1, sum = M[i][j];
for (int k = 0; k < 8; k++) {
int nextX = i + x[k], nextY = j + y[k];
if (nextX >= 0 && nextX < row && nextY >= 0 && nextY < col) {
count++;
sum += M[nextX][nextY];
}
}
return sum / count;
}
}
Python:
class Solution:
def imageSmoother(self, M: List[List[int]]) -> List[List[int]]:
row, col, result = len(M), len(M[0]), [[0] * len(M[0]) for _ in range(len(M))]
M = [[None] * (col + 2)] + [[None] + i + [None] for i in M] + [[None] * (col + 2)]
for i in range(1, row + 1):
for j in range(1, col + 1):
temp = [M[i - 1][j - 1], M[i][j - 1], M[i + 1][j - 1], M[i - 1][j], M[i][j], M[i + 1][j], M[i - 1][j + 1], M[i][j + 1], M[i + 1][j + 1]]
temp = [k for k in temp if k is not None]
result[i - 1][j - 1] = sum(temp) // len(temp)
return result