受Raymond Chen帖子的启发，假设你有一个4x4二维数组，写一个函数使它旋转90度。Raymond链接到伪代码中的解决方案，但我想看到一些真实的东西。
[1][2][3][4]
[5][6][7][8]
[9][0][1][2]
[3][4][5][6]
就变成:
[3][9][5][1]
[4][0][6][2]
[5][1][7][3]
[6][2][8][4]
更新:Nick的答案是最直接的，但是有没有比n²更好的方法呢?如果矩阵是10000x10000呢?
当前回答
试试我图书馆的算盘——常见的:
@Test
public void test_42519() throws Exception {
final IntMatrix matrix = IntMatrix.range(0, 16).reshape(4);
N.println("======= original =======================");
matrix.println();
// print out:
// [0, 1, 2, 3]
// [4, 5, 6, 7]
// [8, 9, 10, 11]
// [12, 13, 14, 15]
N.println("======= rotate 90 ======================");
matrix.rotate90().println();
// print out:
// [12, 8, 4, 0]
// [13, 9, 5, 1]
// [14, 10, 6, 2]
// [15, 11, 7, 3]
N.println("======= rotate 180 =====================");
matrix.rotate180().println();
// print out:
// [15, 14, 13, 12]
// [11, 10, 9, 8]
// [7, 6, 5, 4]
// [3, 2, 1, 0]
N.println("======= rotate 270 ======================");
matrix.rotate270().println();
// print out:
// [3, 7, 11, 15]
// [2, 6, 10, 14]
// [1, 5, 9, 13]
// [0, 4, 8, 12]
N.println("======= transpose =======================");
matrix.transpose().println();
// print out:
// [0, 4, 8, 12]
// [1, 5, 9, 13]
// [2, 6, 10, 14]
// [3, 7, 11, 15]
final IntMatrix bigMatrix = IntMatrix.range(0, 10000_0000).reshape(10000);
// It take about 2 seconds to rotate 10000 X 10000 matrix.
Profiler.run(1, 2, 3, "sequential", () -> bigMatrix.rotate90()).printResult();
// Want faster? Go parallel. 1 second to rotate 10000 X 10000 matrix.
final int[][] a = bigMatrix.array();
final int[][] c = new int[a[0].length][a.length];
final int n = a.length;
final int threadNum = 4;
Profiler.run(1, 2, 3, "parallel", () -> {
IntStream.range(0, n).parallel(threadNum).forEach(i -> {
for (int j = 0; j < n; j++) {
c[i][j] = a[n - j - 1][i];
}
});
}).printResult();
}
其他回答
当前所有的解决方案都有O(n^2)开销作为临时空间(这不包括那些肮脏的OOP骗子!)这里有一个内存占用为O(1)的解决方案，将矩阵原地右转90度。该死的延展性，这玩意儿跑得很快!
#include <algorithm>
#include <cstddef>
// Rotates an NxN matrix of type T 90 degrees to the right.
template <typename T, size_t N>
void rotate_matrix(T (&matrix)[N][N])
{
for(size_t i = 0; i < N; ++i)
for(size_t j = 0; j <= (N-i); ++j)
std::swap(matrix[i][j], matrix[j][i]);
}
免责声明:我实际上并没有测试这个。让我们玩打虫游戏吧!
我的c#示例代码的伟大算法发送@dimple:
/* Author: Dudi,
* http://www.tutorialspoint.com/compile_csharp_online.php?PID=0Bw_CjBb95KQMYm5qU3VjVGNuZFU */
using System.IO;
using System;
class Program
{
static void Main()
{
Console.WriteLine("Rotating this matrix by 90+ degree:");
int[,] values=new int[3,3]{{1,2,3}, {4,5,6}, {7,8,9}};
//int[,] values=new int[4,4]{{101,102,103, 104}, {105,106, 107,108}, {109, 110, 111, 112}, {113, 114, 115, 116}};
print2dArray(ref values);
transpose2dArray(ref values);
//print2dArray(ref values);
reverse2dArray(ref values);
Console.WriteLine("Output:");
print2dArray(ref values);
}
static void print2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen; m++){
Console.Write(matrix[n,m] +"\t");
}
Console.WriteLine();
}
Console.WriteLine();
}
static void transpose2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen; m++){
if(n>m){
int tmp = matrix[n,m];
matrix[n,m] = matrix[m,n];
matrix[m,n] = tmp;
}
}
}
}
static void reverse2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen/2; m++){
int tmp = matrix[n,m];
matrix[n,m] = matrix[n, mLen-1-m];
matrix[n,mLen-1-m] = tmp;
}
}
}
}
/*
Rotating this matrix by 90+ degree:
1 2 3
4 5 6
7 8 9
Output:
7 4 1
8 5 2
9 6 3
*/
在JavaScript中实现dimple的+90伪代码(例如转置然后反转每一行):
function rotate90(a){
// transpose from http://www.codesuck.com/2012/02/transpose-javascript-array-in-one-line.html
a = Object.keys(a[0]).map(function (c) { return a.map(function (r) { return r[c]; }); });
// row reverse
for (i in a){
a[i] = a[i].reverse();
}
return a;
}
PHP:
<?php
$a = array(array(1,2,3,4),array(5,6,7,8),array(9,0,1,2),array(3,4,5,6));
$b = array(); //result
while(count($a)>0)
{
$b[count($a[0])-1][] = array_shift($a[0]);
if (count($a[0])==0)
{
array_shift($a);
}
}
从PHP5.6开始，数组转位可以通过一个狡猾的array_map()调用来执行。换句话说，列被转换为行。
代码:(演示)
$array = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6]
];
$transposed = array_map(null, ...$array);
美元转置:
[
[1, 5, 9, 3],
[2, 6, 0, 4],
[3, 7, 1, 5],
[4, 8, 2, 6]
]
从线性的角度来看，考虑以下矩阵:
1 2 3 0 0 1
A = 4 5 6 B = 0 1 0
7 8 9 1 0 0
现在求A
1 4 7
A' = 2 5 8
3 6 9
考虑A'对B的作用，或B对A'的作用。 分别为:
7 4 1 3 6 9
A'B = 8 5 2 BA' = 2 5 8
9 6 3 1 4 7
这对任何nxn矩阵都是可展开的。 在代码中快速应用这个概念:
void swapInSpace(int** mat, int r1, int c1, int r2, int c2)
{
mat[r1][c1] ^= mat[r2][c2];
mat[r2][c2] ^= mat[r1][c1];
mat[r1][c1] ^= mat[r2][c2];
}
void transpose(int** mat, int size)
{
for (int i = 0; i < size; i++)
{
for (int j = (i + 1); j < size; j++)
{
swapInSpace(mat, i, j, j, i);
}
}
}
void rotate(int** mat, int size)
{
//Get transpose
transpose(mat, size);
//Swap columns
for (int i = 0; i < size / 2; i++)
{
for (int j = 0; j < size; j++)
{
swapInSpace(mat, i, j, size - (i + 1), j);
}
}
}
