Description

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

Solution

Two-pointer, O(mn), S(1)

可以找找规律，很容易就会发现：

• 00 -> 11
• 01 -> 01
• 10 -> 10
• 11 -> 00

所以可以用双指针写，注意双指针重合时（即处理到最中间元素时），别重复flip。

class Solution {
    public int[][] flipAndInvertImage(int[][] A) {
        for (int[] row : A) {
            for (int i = 0, j = row.length - 1; i <= j; ++i, --j) {
                if (row[i] == row[j]) {
                    row[i] ^= 1;
                    if (i < j) {    // avoid flipping twice
                        row[j] ^= 1;
                    }
                }
            }
        }
        
        return A;
    }
}

由于需要特殊处理，所以还不如下面这种general写法：

class Solution {
    public int[][] flipAndInvertImage(int[][] A) {
        for (int[] row : A) {
            for (int i = 0, j = row.length - 1; i <= j; ++i, --j) {
                int tmp = row[i] ^ 1;
                row[i] = row[j] ^ 1;
                row[j] = tmp;
            }
        }
        
        return A;
    }
}