LeetCode/Array/flipAndInvertImage.py at master · konantian/LeetCode · GitHub

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'''
Source : https://leetcode.com/problems/flipping-an-image/
Author : Yuan Wang
Date   : 2018-06-16

/**********************************************************************************
*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]]
*Time complexity:O(n), Space complexity:O(1)
**********************************************************************************/
'''

def convert(x):
	x=1 if x == 0 else 0
	return x
def flipAndInvertImage(A):
	"""
	:type A: List[List[int]]
	:rtype: List[List[int]]
	"""
	for i in A:
		i[:]=i[::-1]
		i[:]=map(convert,i)

	return A

def flipAndInvertImageB(A):
	for row in A:
		for i in xrange((len(row) + 1) / 2):
			row[i], row[~i] = row[~i] ^ 1, row[i] ^ 1
	return A

def flipAndInvertImageC(A):
	"""
	:type A: List[List[int]]
	:rtype: List[List[int]]
	"""

	return [[1-i for i in row[::-1]] for row in A]

A=[[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
print(flipAndInvertImage(A))