README_EN.md

832. Flipping an Image

中文文档

Description

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.

Solutions

Python3

class Solution:
    def flipAndInvertImage(self, A: List[List[int]]) -> List[List[int]]:
        m, n = len(A), len(A[0])
        for i in range(m):
            p, q = 0, n - 1
            while p < q:
                t = A[i][p] ^ 1
                A[i][p] = A[i][q] ^ 1
                A[i][q] = t
                p += 1
                q -= 1
            if p == q:
                A[i][p] ^= 1
        return A

Java

class Solution {
    public int[][] flipAndInvertImage(int[][] A) {
        int m = A.length, n = A[0].length;
        for (int i = 0; i < m; ++i) {
            int p = 0, q = n - 1;
            while (p < q) {
                int t = A[i][p] ^ 1;
                A[i][p] = A[i][q] ^ 1;
                A[i][q] = t;
                ++p;
                --q;
            }
            if (p == q) {
                A[i][p] ^= 1;
            }
        }
        return A;
    }
}

C++

class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
        int m = A.size(), n = A[0].size();
        for (int i = 0; i < m; ++i) {
            int p = 0, q = n - 1;
            while (p < q) {
                int t = A[i][p] ^ 1;
                A[i][p] = A[i][q] ^ 1;
                A[i][q] = t;
                ++p;
                --q;
            }
            if (p == q) {
                A[i][p] ^= 1;
            }
        }
        return A;
    }
};

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