Leetcode 832. Flipping an Image, Java解法

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;
    }