Flipping an Image

題目連結

題目描述

給定一個二維矩陣A。翻轉它，並取反，返回結果。

注：

• 翻轉：顛倒圖片每行。例如[1, 1, 0]變為 [0, 1, 1]。
• 取反：所有的0換成1，所有的1換成0。

約定

1 <= A.length = A[0].length <= 20
0 <= A[i][j] <= 1

示例

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]]
 

解答

解答1

class Solution
{
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A)
    {
        int rowNum = A.size();
        int colNum = A[0].size();
        int colMid = (colNum+1)/2;
        //按行遍歷
        for (int rowIndex = 0; rowIndex < rowNum; rowIndex++)
        {
            for (int colIndex = 0; colIndex < colMid; colIndex++)
            {
                int temp = A[rowIndex][colIndex];
                A[rowIndex][colIndex] = A[rowIndex][colNum - colIndex - 1] == 1 ? 0 : 1;
                A[rowIndex][colNum - colIndex - 1] = temp == 1 ? 0 : 1;
            }
        }
        return A;
    }
};
 

以上程式碼比較簡潔，耗時16ms。

解答2

class Solution
{
public:
    vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A)
    {
        flip(A);
        invert(A);
        return A;
    }
private:
    void flip(vector<vector<int>>& A)
    {
        int rowNum = A.size();
        int colNum = A[0].size();
        //按行遍歷
        for (int rowIndex = 0; rowIndex < rowNum; rowIndex++)
        {
            for (int colIndex = 0; colIndex < colNum/2; colIndex++)
            {
                int temp = A[rowIndex][colIndex];
                A[rowIndex][colIndex] = A[rowIndex][colNum - colIndex - 1];
                A[rowIndex][colNum - colIndex - 1] = temp;
            }
        }
    }
    void invert(vector<vector<int>>& A)
    {
        int rowNum = A.size();
        int colNum = A[0].size();
        for (int rowIndex = 0; rowIndex < rowNum; rowIndex++)
        {
            for (int colIndex = 0; colIndex < colNum; colIndex++)
            {
                A[rowIndex][colIndex] = A[rowIndex][colIndex] == 1 ? 0 : 1;
            }
        }
    }
};
 

此種方法不太簡潔，但耗時8ms，超過99.5%。