[LeetCode] 832번 - Flipping an Image(상 뒤집기)

문제

832. Flipping an Image

Easy

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

풀이과정

(1)규칙

• 배열을 반대로 정렬합니다( [0,1,1]일 경우, [1,1,0] )
• 각 요소의 0은 1로, 1은 0으로 바꿔줍니다

(2)순서

swap()으로 배열요소를 반대로 정렬하고, 모든값에 XOR을 한번씩 해주거나,(반복문 2번 사용)

배열길이/2까지만 순회하며 XOR한 값들을 swap해줍니다.(반복문1번 사용)

(3)코드

C++ / 1번 방법

class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
        for (int k = 0; k < A.size(); k++) {
            int i = 0, j = A.front().size() - 1;
            while (i < j) swap(A[k][i++], A[k][j--]); //reverse
            for (i = 0; i < A.front().size(); i++) A[k][i]^=1; //flip
        }
        return A;
    }
};

C++ / 2번 방법

class Solution {
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
        for (int i = 0; i < A.size(); i++) {
            for(int j = 0 ; j < (A.front().size()+1)/2;j++){
                int temp = A[i][j]^1;
                A[i][j]=A[i][A.front().size()-1-j]^1;
                A[i][A.front().size()-1-j]=temp;
            }
        }
        return A;
    }
};