Leetcode 832. Flipping an Image, Java解法

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博客围绕二进制矩阵展开，要对其进行水平翻转和反转操作。介绍了两种解法，一是用for循环结合Stack进行反转，二是从两头开始比较，第二种方法更节省时间。

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

Approach #1 我的解法, for循环结合Stack进行reverse

class Solution {
    public int[][] flipAndInvertImage(int[][] A) {
       
        int len = A[0].length;
        for(int i=0;i<len;i++){
            Stack<Integer> st = new Stack<Integer>();
            for(int j=0;j<len;j++){
                st.push(A[i][j]);
            }
            
            for(int j=0;j<len;j++){
                if(st.pop() == 0) A[i][j] =1;
                else A[i][j] = 0;
            }
        }
        return A;
    }
}

Approach #2 从两头开始比较，更节省时间

public int[][] flipAndInvertImage(int[][] A) {
				
        for(int k=0; k<A.length; k++)
        {
        	int i=0;
        	int j=A[k].length-1;
        	while(i<=j)
        	{
        		int temp = A[k][i];
        		A[k][i] = (A[k][j] > 0) ? 0 : 1;
        		A[k][j] = (temp > 0) ? 0 : 1;
        		i++;
        		j--;
        	}        	
        }
        
        return A;
    }