如何将 N x N 矩阵旋转90度,且需要原地完成?
如何将 N x N 矩阵旋转90度,且需要原地完成?
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;
}
注意 我没有测试过这个代码,只是现场编写的。在实际使用前,请务必进行测试。
这是我的解决方案: (顺时针旋转90度)
对数组进行转置(类似于矩阵转置)
反转每一行的元素
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])
}
}
如果逆时针旋转π/2
转置数组
反转列顺序的元素
不要测试代码! 欢迎任何建议!
//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);
}
}
#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;
}
你可以创建第二个数组,然后通过在第一个数组中按行主序读取并在第二个数组中按列主序写入来将第一个数组复制到第二个数组中。
因此,你需要复制:
1 2 3
4 5 6
7 8 9
而你需要读取第一行,然后从那里开始向上写:
3
2
1