1、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

代码

 public int[][] flipAndInvertImage(int[][] A) {
        for(int i=0;i<A.length;i++) {
            int[] new_array = new int[A[i].length]; 
            for(int j=0;j<A[i].length;j++) {
                new_array[j]= A[i][A[i].length-j-1]^1;
            }
            A[i] = new_array;
        }
        return A;
    }

注意：异或运算符是用符号“^”表示的，其运算规律是：
两个操作数的位中，相同则结果为0，不同则结果为1

2、Sort Array By Parity

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

1 <= A.length <= 5000
0 <= A[i] <= 5000

代码

    public int[] sortArrayByParity(int[] A) {
        
        int i = 0;
        int j = A.length-1;
        
        while(i<j){         
                if(A[i]%2!=0&&A[j]%2==0){
                    int tmp = A[j];
                    A[j] = A[i];
                    A[i] = tmp;
                }
                
                if(A[i]%2==0) i++;   
                if(A[j]%2!=0)  j--;                      
        }
         
        return A;
    }