LeetCode 832. Flipping an Image

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

输入一个矩阵，先对每一行进行左右反转、再将每一个值由0变为1,1变为0

public class Solution {

    public static int[][] flipAndInvertImage(int[][] A) {
        int half = A[0].length >> 1;
        half = (A[0].length & 1) == 0 ? half-1:half;
        for(int i = 0; i < A.length; i++) {
            for(int j = 0; j <= half; j++) {
                int temp = A[i][j];
                A[i][j] = 1 - A[i][A[i].length-j-1];
                A[i][A[i].length-j-1] = 1 - temp;
            }
        }
        return A;
    }


    public static void main(String[] args) {
//        int[][] A = {{1}};
//        int[][] A = {{1,1,0},{1,0,1},{0,0,0}};
        int[][] A = {{1,1,0,0},{1,0,0,1},{0,1,1,1},{1,0,1,0}};
        int[][] r = Solution.flipAndInvertImage(A);
        for(int i = 0; i < r.length; i++) {
            for(int j = 0; j < r[i].length; j++) {
                System.out.print(r[i][j]+" ");
            }
            System.out.println();
        }
    }
}