[LeetCode] 832. Flipping an Image_Easy

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

基本思路, 先flip每行, 然后翻转. T: O(m*n)  S; O(m*n)

Improve: 可以update in place, for i in range(len(row)//2), 然后将left和right翻转之后对调, 即可, 但是时间复杂度一样 O(m*n)

Code

class Solution:
    def flipImage(self, A):
        lr, lc, ans = len(A), len(A[0]), A
        for i in range(lr):
            ans[i] = A[i][::-1]
        for i in range(lr):
            for j in range(lc):
                ans[i][j] = 1- ans[i][j]
        return ans