Leetcode 832 Flipping an Image

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;
    }