LeetCode 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

题目描述：将一个 n*n 的矩阵进行水平翻转以后再进行图像反转( 0 变 1 , 1 变 0 )，求解此时反转后的矩阵。

题目分析：其实很简单，我们将这个过程分为 2 步，第一步是水平翻转，第二步是图像反转，水平翻转我们可以利用 reverse 函数或者申请一个同样规模的矩阵去水平逆序存储矩阵值，图像反转很容易，通过遍历查找，将 1 替换成 0 ， 0 替换成 1 即可。

python 代码：

class Solution(object):
    def flipAndInvertImage(self, A):
        """
        :type A: List[List[int]]
        :rtype: List[List[int]]
        """
        A_length = len(A)
        A_length_index = len(A[A_length-1])
        for i in range(A_length):
            A[i].reverse()
            
        for i in range(A_length):
            for j in range(A_length_index):
                if A[i][j] == 0:
                    A[i][j] = 1
                else:
                    A[i][j] = 0
        return A

C++ 代码：

class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
        vector<vector<int>> B(A.size());
        //vector<int>::reverse_iterator r_iter;
        for(int i = 0; i < B.size(); i++){
            B[i].resize(A[0].size());
        }
        
        for(int i = 0; i < A.size(); i++){
            int x = 0;
            for(int j = A[i].size()-1; j >= 0; j--){
                B[i][x] = A[i][j];
                x = x + 1;
            }
        }
        
        for(int i = 0; i < B.size(); i++){
            for(int j = 0; j < B[i].size(); j++){
                if(B[i][j] == 1){
                    B[i][j] = 0;
                }
                else{
                    B[i][j] = 1;
                }
            }
        } 
        return B;
    }
};