# 832. Flipping an Image 翻转图像

@TOC

## # 题目描述

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. 1 <= A.length = A[0].length <= 20
2. 0 <= A[i][j] <= 1

## # 解题方法

### # 翻转 + 异或

``````class Solution:
def flipAndInvertImage(self, A):
"""
:type A: List[List[int]]
:rtype: List[List[int]]
"""
rows = len(A)
cols = len(A[0])
for row in range(rows):
A[row] = A[row][::-1]
for col in range(cols):
A[row][col] ^= 1
return A
``````

### # 直接计算

``````class Solution:
def flipAndInvertImage(self, A):
"""
:type A: List[List[int]]
:rtype: List[List[int]]
"""
M, N = len(A), len(A[0])
res = [[0] * N for _ in range(M)]
for i in range(M):
for j in range(N):
res[i][j] = 1 - A[i][N - 1 - j]
return res
``````

## # 日期

2018 年 5 月 27 日 —— 很久没刷题了，这个是恢复的第一个题目。 2018 年 11 月 3 日 —— 雾霾的周六