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。
1 #include <iostream> 2 #include <fstream> 3 #include <cstring> 4 #include <vector> 5 #include <assert.h> 6 7 using namespace std; 8 9 void rotate( vector<vector<unsigned int> >& image ); 10 11 int main( int argc, char *argv[] ) { 12 string data_file = "./1.6.txt"; 13 ifstream ifile( data_file.c_str(), ios::in ); 14 if( !ifile.is_open() ) { 15 fprintf( stderr, "cannot open file: %s ", data_file.c_str() ); 16 return -1; 17 } 18 int n = 0; 19 while( ifile >>n ) { 20 vector<vector<unsigned int> > image( n, vector<unsigned int>( n, 0 ) ); 21 assert( n > 0 ); 22 cout <<"input:" <<endl; 23 for( int i = 0; i < n; ++i ) { 24 for( int j = 0; j < n; ++j ) { 25 ifile >>image[i][j]; 26 cout <<image[i][j] <<" "; 27 } 28 cout <<endl; 29 } 30 rotate( image ); 31 cout <<"result:" <<endl; 32 for( int i = 0; i < n; ++i ) { 33 for( int j = 0; j < n; ++j ) { 34 cout <<image[i][j] <<" "; 35 } 36 cout <<endl; 37 } 38 cout <<endl; 39 } 40 ifile.close(); 41 return 0; 42 } 43 44 void rotate( vector<vector<unsigned int> >& image ) { 45 int n = image.size(); 46 if( --n <= 0 ) { return; } 47 for( int i = 0; i <= n/2; ++i ) { 48 for( int j = i; j < n-i; ++j ) { 49 unsigned int tmp = image[i][j]; 50 image[i][j] = image[n-j][i]; 51 image[n-j][i] = image[n-i][n-j]; 52 image[n-i][n-j] = image[j][n-i]; 53 image[j][n-i] = tmp; 54 } 55 } 56 return; 57 }
测试文件
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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