[LeetCode/Scala] 832. Flipping an Image

最新推荐文章于 2021-03-01 20:37:35 发布

原创最新推荐文章于 2021-03-01 20:37:35 发布 · 131 阅读

CC 4.0 BY-SA版权

本文介绍了一种高效的算法，用于处理二进制矩阵的水平翻转和元素反转。通过使用map函数和位运算，避免了传统的for循环，实现了对矩阵的快速处理。示例展示了如何将输入矩阵[[1,1,0],[1,0,1],[0,0,0]]转换为[[1,0,0],[0,1,0],[1,1,1]]。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary matrix A, we want to flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed. For example, flipping [1, 1, 0] horizontally results in [0, 1, 1].

To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0. For example, inverting [0, 1, 1] results in [1, 0, 0].

Example 1:

Input: [[1,1,0],[1,0,1],[0,0,0]]
Output: [[1,0,0],[0,1,0],[1,1,1]]
Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]
Example 2:

Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Notes:

1 <= A.length = A[0].length <= 20
0 <= A[i][j] <= 1

用map函数
$_\_$ 符号应用
算法对 $A$ 中的每个元素 $x$ ，进行 $r e v e r s e$ 操作，对得到结果进行 $f l i p$ 操作。
不在需要 $loopfor\ loop$

object Solution {
    def flipAndInvertImage(A: Array[Array[Int]]): Array[Array[Int]] =
    A map (_.reverse map (_^1))
}