LeetCode 832. Flipping an Image-Toy模板网

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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.

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。文章来源地址https://www.toymoban.com/news/detail-675317.html

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;
    }
}

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