LeetCode-in-Java

832. Flipping an Image

Easy

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.

Solution

public class Solution {
    public int[][] flipAndInvertImage(int[][] image) {
        int m = image.length;
        int n = image[0].length;
        int[][] result = new int[m][n];
        for (int i = 0; i < m; i++) {
            int[] flipped = (reverse(image[i]));
            result[i] = invert(flipped);
        }
        return result;
    }

    private int[] invert(int[] flipped) {
        int[] result = new int[flipped.length];
        for (int i = 0; i < flipped.length; i++) {
            if (flipped[i] == 0) {
                result[i] = 1;
            } else {
                result[i] = 0;
            }
        }
        return result;
    }

    private int[] reverse(int[] nums) {
        for (int i = 0, j = nums.length - 1; i < j; i++, j--) {
            int tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
        }
        return nums;
    }
}

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