数组与字符串 1.6

给定一幅NXN矩阵表示的图像，其中每个像素的大小为4字节，编写一个方法，将图像旋转90度。不占用额外内存空间能否做到？

分析：此处假设对图像做顺时针旋转。对于image[i][j]，其顺时针旋转角度与对应点坐标分别为：90度--image[j][n-i-1], 180度--image[n-i][n-j], 270度--image[n-j][i]。赋值流程为：tmp <-- image[i][j] <-- image[n-i-1][j] <-- image[n-i-1][n-j-1] <-- image[i][n-j-1] <-- tmp。

#include <iostream>
#include <fstream>
#include <cstring>
#include <vector>
#include <assert.h>

using namespace std;

void rotate( vector<vector<unsigned int> >& image );

int main( int argc, char *argv[] ) {
    string data_file = "./1.6.txt";
    ifstream ifile( data_file.c_str(), ios::in );
    if( !ifile.is_open() ) {
        fprintf( stderr, "cannot open file: %s
", data_file.c_str() );
        return -1;
    }
    int n = 0;
    while( ifile >>n ) {
        vector<vector<unsigned int> > image( n, vector<unsigned int>( n, 0 ) );
        assert( n > 0 );
        cout <<"input:" <<endl;
        for( int i = 0; i < n; ++i ) {
            for( int j = 0; j < n; ++j ) {
                ifile >>image[i][j];
                cout <<image[i][j] <<" ";
            }
            cout <<endl;
        }
        rotate( image );
        cout <<"result:" <<endl;
        for( int i = 0; i < n; ++i ) {
            for( int j = 0; j < n; ++j ) {
                cout <<image[i][j] <<" ";
            }
            cout <<endl;
        }
        cout <<endl;
    }
    ifile.close();
    return 0;
}

void rotate( vector<vector<unsigned int> >& image ) {
    int n = image.size();
    if( --n <= 0 ) { return; }
    for( int i = 0; i <= n/2; ++i ) {
        for( int j = i; j < n-i; ++j ) {
            unsigned int tmp = image[i][j];
            image[i][j] = image[n-j][i];
            image[n-j][i] = image[n-i][n-j];
            image[n-i][n-j] = image[j][n-i];
            image[j][n-i] = tmp;
        }
    }
    return;
}

测试文件

1
1

2
1 2
3 4

3
1 2 3
4 5 6
7 8 9

4
1 2 3 4
4 5 6 7
9 10 11 12
13 14 15 16

5
232 123 243 231 211
654 123 123 134 876
142 562 475 987 321
235 635 478 693 146
111 222 323 454 656