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
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!
其他回答
这是一个如今被高估的面试问题。
我的建议是:不要让面试官用他们关于解决这个问题的疯狂建议把你弄糊涂了。使用白板绘制输入数组的索引，然后绘制输出数组的索引。旋转前后的列分度示例如下:
30 --> 00
20 --> 01
10 --> 02
00 --> 03
31 --> 10
21 --> 11
11 --> 12
01 --> 13
注意旋转后的数字模式。
下面提供了一个简洁的Java解决方案。经过测试，它是有效的:
Input:
M A C P
B N L D
Y E T S
I W R Z
Output:
I Y B M
W E N A
R T L C
Z S D P
/**
* (c) @author "G A N MOHIM"
* Oct 3, 2015
* RotateArrayNintyDegree.java
*/
package rotatearray;
public class RotateArrayNintyDegree {
public char[][] rotateArrayNinetyDegree(char[][] input) {
int k; // k is used to generate index for output array
char[][] output = new char[input.length] [input[0].length];
for (int i = 0; i < input.length; i++) {
k = 0;
for (int j = input.length-1; j >= 0; j--) {
output[i][k] = input[j][i]; // note how i is used as column index, and j as row
k++;
}
}
return output;
}
public void printArray(char[][] charArray) {
for (int i = 0; i < charArray.length; i++) {
for (int j = 0; j < charArray[0].length; j++) {
System.out.print(charArray[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
char[][] input =
{ {'M', 'A', 'C', 'P'},
{'B', 'N', 'L', 'D'},
{'Y', 'E', 'T', 'S'},
{'I', 'W', 'R', 'Z'}
};
char[][] output = new char[input.length] [input[0].length];
RotateArrayNintyDegree rotationObj = new RotateArrayNintyDegree();
rotationObj.printArray(input);
System.out.println("\n");
output = rotationObj.rotateArrayNinetyDegree(input);
rotationObj.printArray(output);
}
}
这是将数组旋转90度的简单C代码。希望这能有所帮助。
#include <stdio.h>
void main(){
int arr[3][4] = {85, 2, 85, 4,
85, 6, 7, 85,
9, 85, 11, 12};
int arr1[4][3];
int i = 0, j = 0;
for(i=0;i<4;i++){
int k = 2;//k = (number of columns in the new array arr1 - 1)
for(j=0;j<3;j++){
arr1[i][j]=arr[k][i];
k--;
}
}
int l, m;
for(l=0;l<4;l++){
for(m=0;m<3;m++){
printf("%d ", arr1[l][m]);
}
printf("\n");
}
}//end main
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
*/
JavaScript解决方案旋转矩阵90度的地方:
function rotateBy90(m) {
var length = m.length;
//for each layer of the matrix
for (var first = 0; first < length >> 1; first++) {
var last = length - 1 - first;
for (var i = first; i < last; i++) {
var top = m[first][i]; //store top
m[first][i] = m[last - i][first]; //top = left
m[last - i][first] = m[last][last - i]; //left = bottom
m[last][last - i] = m[i][last]; //bottom = right
m[i][last] = top; //right = top
}
}
return m;
}
Javascript解决NxN矩阵与运行时O(N^2)和内存O(1)
function rotate90(matrix){
var length = matrix.length
for(var row = 0; row < (length / 2); row++){
for(var col = row; col < ( length - 1 - row); col++){
var tmpVal = matrix[row][col];
for(var i = 0; i < 4; i++){
var rowSwap = col;
var colSwap = (length - 1) - row;
var poppedVal = matrix[rowSwap][colSwap];
matrix[rowSwap][colSwap] = tmpVal;
tmpVal = poppedVal;
col = colSwap;
row = rowSwap;
}
}
}
}
