LeetCode 832. Flipping an Image

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修改于 2018-10-29 17:13:43

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修改于 2018-10-29 17:13:43

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832. Flipping an Image

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].

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:

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

解法：

//832. Flipping an Image
public int[][] flipAndInvertImage(int[][] A) {
    if (A==null||A.length== 0||A[0].length == 0) return A;
    for (int j = 0;j<A.length;j++){
        for (int i = 0;i<A[0].length/2;i++){
            int temp = A[j][i];
            A[j][i] = A[j][A[0].length-1-i];
            A[j][A[0].length-1-i] = temp;
        }
    }
    for (int i = 0;i<A.length;i++){
        for (int j = 0;j<A[0].length;j++){
            if (A[i][j] ==1) A[i][j] =0;
            else if (A[i][j] ==0) A[i][j] =1;
        }
    }
    return A;
}

题目简单直接暴力解法就完事了。

原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。

如有侵权，请联系 cloudcommunity@tencent.com 删除。

其他

原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。

如有侵权，请联系 cloudcommunity@tencent.com 删除。

其他

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