0832._Flipping_an_Image.md

832. Flipping an Image

难度: Easy

刷题内容

原题连接

• https://leetcode.com/problems/flipping-an-image/

内容描述

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

解题方案

思路 1 - 时间复杂度: O(row * col)- 空间复杂度: O(1)******

直接in-place改, beats 38.76%

class Solution(object):
    def flipAndInvertImage(self, A):
        """
        :type A: List[List[int]]
        :rtype: List[List[int]]
        """
        for i in range(len(A)):
            for j in range((len(A[0])+1)//2):
                A[i][j], A[i][-j-1] = 0 if A[i][-j-1] else 1, 0 if A[i][j] else 1
        return A