9/11 每日一題(對一個nxn矩陣 先進行水平翻轉, 再進行每個元素的0、1翻轉)

 Given an n x n binary matrix image, 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: image = [[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: image = [[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]]

 

Constraints:

• n == image.length
• n == image[i].length
• 1 <= n <= 20
• images[i][j] is either 0 or 1.

class Solution:

    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:

        for i in range(len(image)):

            image[i]=image[i][::-1] #水平翻轉

       

        for i in image:

            n=0

            for j in i: #將1、0翻轉

                i[n]=1-j

                n+=1

        return image

               

       

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class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        ans = []
        for im in image:
            flip = im[::-1]
            invert =[]
            for num in flip:
                if num == 0:
                    invert.append(1)
                else:
                    invert.append(0)
            ans.append(invert)
        return ans