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Improve 开发者_JS百科this questionHow to rotate a N x N matrix by 90 degrees. I want it to be inplace?
for(int i=0; i<n/2; i++)
for(int j=0; j<(n+1)/2; j++)
cyclic_roll(m[i][j], m[n-1-j][i], m[n-1-i][n-1-j], m[j][n-1-i]);
void cyclic_roll(int &a, int &b, int &c, int &d)
{
int temp = a;
a = b;
b = c;
c = d;
d = temp;
}
Note I haven't tested this, just compoosed now on the spot. Please test before doing anything with it.
here is my solution: (rotate pi/2 clockwise)
do the transpose of the array, (like matrix transpose)
reverse the elements for each row
cons int row = 10; cons int col = 10; //transpose for(int r = 0; r < row; r++) { for(int c = r; c < col; c++) { swap(Array[r][c], Array[c][r]); } } //reverse elements on row order for(int r = 0; r < row; r++) { for(int c =0; c < col/2; c++) { swap(Array[r][c], Array[r][col-c-1]) } }
if rotate pi/2 in counter-clockwise
transpose the array
reverse the elements on column order
never test the code! any suggestion would be appreciated!
A complete C program which illustrates my approach. Essentially it's recursive algo. At each recursion you rotate the outer layer. Stop when your matrix is 1x1 or 0x0.
#include <stdio.h>
int matrix[4][4] = {
{11, 12, 13, 14},
{21, 22, 23, 24},
{31, 32, 33, 34},
{41, 42, 43, 44}
};
void print_matrix(int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf(" %d ", matrix[i][j]);
}
printf("\n");
}
}
int *get(int offset, int x, int y)
{
return &matrix[offset + x][offset + y];
}
void transpose(int offset, int n)
{
if (n > 1) {
for (int i = 0; i < n - 1; i++) {
int *val1 = get(offset, 0, i);
int *val2 = get(offset, i, n - 1);
int *val3 = get(offset, n - 1, n - 1 - i);
int *val4 = get(offset, n - 1 - i, 0);
int temp = *val1;
*val1 = *val4;
*val4 = *val3;
*val3 = *val2;
*val2 = temp;
}
transpose(offset + 1, n - 2);
}
}
main(int argc, char *argv[])
{
print_matrix(4);
transpose(0, 4);
print_matrix(4);
return 0;
}
//Java version, fully tested
public class Rotate90degree {
public static void reverseElementsRowWise(int[][] matrix) {
int n = matrix.length;
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n / 2; ++j) {
int temp = matrix[i][n - j - 1];
matrix[i][n - j - 1] = matrix[i][j];
matrix[i][j] = temp;
}
}
}
public static void transpose(int[][] matrix) {
int n = matrix.length;
for(int i = 0; i < n; ++i) {
for(int j = i + 1; j < n; ++j) {
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
}
public static void rotate90(int[][] matrix) {
transpose(matrix);
reverseElementsRowWise(matrix);
}
public static void print(int[][] matrix) {
int n = matrix.length;
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n; ++j) {
System.out.print(matrix[i][j]);
System.out.print(' ');
}
System.out.println();
}
}
public static void main(String[] args) {
int[][] matrix = {
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}};
System.out.println("before");
print(matrix);
rotate90(matrix);
System.out.println("after");
print(matrix);
}
}
You could create a second array and then copy the first one into the second one by reading row-major in the first one and writing column-major to the second one.
So you would copy:
1 2 3
4 5 6
7 8 9
and you would read the first row then write it back up starting like:
3
2
1
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