Leetcode 832 Flipping an Image

最新推荐文章于 2021-02-09 22:27:17 发布

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博客围绕二进制矩阵处理问题展开，题目要求先将矩阵图像水平翻转，再进行反转（0变1，1变0）。给出了解题思路，即先左右翻转，再进行0和1的互换，还给出了示例及相关条件说明。

题目描述

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].

简而言之：先左右翻转，再将0，1互换。

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:（这个条件可以让我们不必判断A.length<0的情况）

1 <= A.length = A[0].length <= 20
0 <= A[i][j] <= 1

解题思路

先左右翻转，再将0置为1,1置为0.

题解

    public int[][] flipAndInvertImage(int[][] A) {
         // 1 <= A.length = A[0].length <= 20 不用判断输入问题
        int n = A.length;
        // flip
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= (n-1)/2; j++) {
                int temp = A[i][j];
                A[i][j] = A[i][n-1-j];
                A[i][n-1-j] = temp;

            }
        }
        // invert
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (A[i][j] == 0) A[i][j] = 1;
                else A[i][j] = 0;
            }
        }
        return A;
    }