Cracking the coding interview 题目

Cracking the coding interview  题目


Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?


译文:


一张图像表示成NxN的矩阵,图像中每个像素是4个字节,写一个函数把图像旋转90度。 你能原地进行操作吗?(即不开辟额外的存储空间)

解决思路比较简单:
首先将整个图当做一个 int型的矩阵,这样,首先将这个整数矩阵沿着右上左下方向的对角线进行翻转,在对整个矩阵进行上下翻转即可得到想要的结果。

#include <iostream>

using namespace std;

#define N 4

int main()
{
    int array[N][N];

    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
            array[i][j] = i * N + j + 1;
        }
    }
    cout << "反转前:" << endl;
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
            cout << array[i][j] << "\t";
        }
        cout << endl;
    }
    cout << endl;
    cout << endl;

    // 先进行对角线翻转
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < i; j++)
        {
            array[i][j] ^= array[j][i];
            array[j][i] ^= array[i][j];
            array[i][j] ^= array[j][i];
        }
    }

    // 再进行上下翻转
    for(int i = 0; i < N / 2; i++)
    {
        for(int j = 0; j < N; j++)
        {
            array[i][j] ^= array[N-1-i][j];
            array[N-1-i][j] ^= array[i][j];
            array[i][j] ^= array[N-1-i][j];
        }
    }
    cout << "反转后:" << endl;
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
            cout << array[i][j] << "\t";
        }
        cout << endl;
    }

    return 0;
}







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