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.
[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
.
[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
.这题要把一个0/1的2d array先横着倒过来,再把每个元素的0/1反过来,其实很直观很简单,就算是in place也很简单,于是就很快写出了代码。我的思路就是对每一行都用2 pointers指向前后,用一个temp来swap就行了,赋值的时候顺便取反。然后看了solutions里大家的思路,太强了,可以不需要temp直接判断前后两个数字一不一样,不一样啥也不用干,一样的话各自取反就行了。太妙了。
class Solution {
public int[][] flipAndInvertImage(int[][] image) {
int row = image.length;
int col = image[0].length;
for (int i = 0; i < row; i++) {
int start = 0;
int end = col - 1;
while (start <= end) {
int temp = image[i][start];
image[i][start] = image[i][end] == 0 ? 1 : 0;
image[i][end] = temp == 0 ? 1 : 0;
start++;
end--;
}
}
return image;
}
}