问题：给定一个整数n，打印数字从1到n 2 ，如下所示：
n = 4
结果是：
01 02 03 04
12 13 14 05
11 16 15 06
10 09 08 07
你如何解决它（除了以下链接中提供的解决方案）？
http://www.programmersheaven.com/mb/CandCPP/81986/81986/problem-in-making-ap-c++-program/?S=B20000
我正朝另一个方向看。到目前为止，我正在试图找出是否可以获得我必须填写的有序位置列表。
以下是我正在研究的内容：有没有办法获得“fdisp”以便解决问题，而不是在矩阵中“走”？
matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
n = len(matrix)
# final disposition wrote by hand: how to get it for arbitrary n?
fdisp = [(0,0), (0,1), (0,2), (0,3), (1,3), (2,3), (3,3), (3,2),
(3,1), (3,0), (2,0), (1,0), (1,1), (1,2), (2,2), (2,1)]
for val,i in enumerate(fdisp):
matrix[i[0]][i[1]] = val + 1
def show_matrix(matrix, n):
for i,l in enumerate(matrix):
for j in range(n):
print "%d\t" % matrix[i][j],
print
show_matrix(matrix, n)
答案 0 :(得分：5)
这是一种不同的方法。它依赖于发现你所做的动作在右，下，左，上，右，......之间循环。此外，你移动的次数：3右，3下，3左，2上，右2 ，1倒，1左。所以不用多说，我会用Python编写代码。
首先，我将使用一些itertools和一些numpy：
from itertools import chain, cycle, imap, izip, repeat
from numpy import array
方向之间的循环：右，下，左，上，右，......：
directions = cycle(array(v) for v in ((0,1),(1,0),(0,-1),(-1,0)))
（我在这里使用numpy的数组，所以我可以很容易地一起添加方向。元组不能很好地添加。）
接下来，我移动的次数从n-1减少到1，重复每个数字两次，第一个数字重复三次：
countdown = chain((n-1,), *imap(repeat, range(n-1,0,-1), repeat(2)))
所以现在可以通过在倒计时中按配对数字重复每个连续方向来创建我的方向序列：
dirseq = chain(*imap(repeat, directions, countdown))
为了得到我的索引序列，我可以总结这个序列，但是（AFAIK）Python没有提供这样的方法，所以让我们快速将它们放在一起：
def sumseq(seq, start=0):
v = start
yield v
for s in seq:
v += s
yield v
现在要生成原始数组，我可以执行以下操作：
a = array(((0,)*n,)*n) # n-by-n array of zeroes
for i, v in enumerate(sumseq(dirseq, array((0,0)))):
a[v[0], v[1]] = i+1
print a
对于n = 4，给出：
[[ 1 2 3 4]
[12 13 14 5]
[11 16 15 6]
[10 9 8 7]]
并且，对于n = 5，给出：
[[ 1 2 3 4 5]
[16 17 18 19 6]
[15 24 25 20 7]
[14 23 22 21 8]
[13 12 11 10 9]]
这种方法可以推广到矩形网格;我将此作为练习留给读者;）
答案 1 :(得分：2)
虽然您的示例是在python中并且这是Java中的，但我认为您应该能够遵循逻辑：
public class SquareTest {
public static void main(String[] args) {
SquareTest squareTest = new SquareTest(4);
System.out.println(squareTest);
}
private int squareSize;
private int[][] numberSquare;
private int currentX;
private int currentY;
private Direction currentDirection;
private enum Direction {
LEFT_TO_RIGHT, RIGHT_TO_LEFT, TOP_TO_BOTTOM, BOTTOM_TO_TOP;
};
public SquareTest(int squareSize) {
this.squareSize = squareSize;
numberSquare = new int[squareSize][squareSize];
currentY = 0;
currentX = 0;
currentDirection = Direction.LEFT_TO_RIGHT;
constructSquare();
}
private void constructSquare() {
for (int i = 0; i < squareSize * squareSize; i = i + 1) {
numberSquare[currentY][currentX] = i + 1;
if (Direction.LEFT_TO_RIGHT.equals(currentDirection)) {
travelLeftToRight();
} else if (Direction.RIGHT_TO_LEFT.equals(currentDirection)) {
travelRightToLeft();
} else if (Direction.TOP_TO_BOTTOM.equals(currentDirection)) {
travelTopToBottom();
} else {
travelBottomToTop();
}
}
}
private void travelLeftToRight() {
if (currentX + 1 == squareSize || numberSquare[currentY][currentX + 1] != 0) {
currentY = currentY + 1;
currentDirection = Direction.TOP_TO_BOTTOM;
} else {
currentX = currentX + 1;
}
}
private void travelRightToLeft() {
if (currentX - 1 < 0 || numberSquare[currentY][currentX - 1] != 0) {
currentY = currentY - 1;
currentDirection = Direction.BOTTOM_TO_TOP;
} else {
currentX = currentX - 1;
}
}
private void travelTopToBottom() {
if (currentY + 1 == squareSize || numberSquare[currentY + 1][currentX] != 0) {
currentX = currentX - 1;
currentDirection = Direction.RIGHT_TO_LEFT;
} else {
currentY = currentY + 1;
}
}
private void travelBottomToTop() {
if (currentY - 1 < 0 || numberSquare[currentY - 1][currentX] != 0) {
currentX = currentX + 1;
currentDirection = Direction.LEFT_TO_RIGHT;
} else {
currentY = currentY - 1;
}
}
@Override
public String toString() {
StringBuilder builder = new StringBuilder();
for (int i = 0; i < squareSize; i = i + 1) {
for (int j = 0; j < squareSize; j = j + 1) {
builder.append(numberSquare[i][j]);
builder.append(" ");
}
builder.append("\n");
}
return builder.toString();
}
}
答案 2 :(得分：2)
另一种方法，这次是在C＃：
int number = 9;
var position = new { x = -1, y = 0 };
var directions = new [] {
new { x = 1, y = 0 },
new { x = 0, y = 1 },
new { x = -1, y = 0 },
new { x = 0, y = -1 }
};
var sequence = (
from n in Enumerable.Range(1, number)
from o in Enumerable.Repeat(n, n != number ? 2 : 1)
select o
).Reverse().ToList();
var result = new int[number,number];
for (int i = 0, current = 1; i < sequence.Count; i++)
{
var direction = directions[i % directions.Length];
for (int j = 0; j < sequence[i]; j++, current++)
{
position = new {
x = position.x + direction.x,
y = position.y + direction.y
};
result[position.y, position.x] = current;
}
}
答案 3 :(得分：1)
dim = n
pos = (0, -1)
fdisp = []
squares = n % 2 == 0 and n / 2 or n / 2 + 1
for _ in range(squares):
pos = (pos[0], pos[1] + 1)
fdisp.append(pos)
fdisp += [(pos[0],pos[1]+i) for i in range(1, dim)]
pos = fdisp[-1]
fdisp += [(pos[0]+i,pos[1]) for i in range(1, dim)]
pos = fdisp[-1]
fdisp += [(pos[0],pos[1]-i) for i in range(1, dim)]
pos = fdisp[-1]
fdisp += [(pos[0]-i,pos[1]) for i in range(1, dim - 1)]
pos = fdisp[-1]
dim = dim - 2
matrix = [[0] * n for i in range(n)]
for val,i in enumerate(fdisp):
matrix[i[0]][i[1]] = val + 1
def show_matrix(matrix, n):
for i,l in enumerate(matrix):
for j in range(n):
print "%d\t" % matrix[i][j],
print
show_matrix(matrix, n)
答案 4 :(得分：1)
我用C ++解决了你的问题。我不知道它对你有帮助。但张贴它。如果它对你有用，那将是一种乐趣。
以下是代码：
#include<iostream>
#include<string.h>
using namespace std;
bool valid(int n,int r,int c)
{
if(r>=1 && r<=n && c>=1 && c<=n)
return true;
return false;
}
int main()
{
pair<int,int>d1,d2,d3,d4,temp;
d1 = make_pair(0,1);
d2 = make_pair(1,0);
d3 = make_pair(0,-1);
d4 = make_pair(-1,0);
/**********************direction******************************/
int n, i, j, counter=1, newR = 1, newC = 0, direction = 4;
bool changeDir=true;
/**************************variables*************************/
cin>>n;
int arr[n+1][n+1];
int visited[n+1][n+1];
/*************************arrays********************************/
memset(visited,0,sizeof(visited));
memset(arr,0,sizeof(arr));
/***************initializing the array**************************/
while(counter<=n*n)
{
if(direction==1 && changeDir)
{
temp = make_pair(d2.first,d2.second);
direction=2;
changeDir=false;
}
else if(direction==2&& changeDir)
{
temp = make_pair(d3.first,d3.second);
direction=3;
changeDir=false;
}
else if(direction==3&& changeDir)
{
temp = make_pair(d4.first,d4.second);
direction=4;
changeDir=false;
}
else if(direction==4&& changeDir)
{
temp = make_pair(d1.first,d1.second);
direction=1;
changeDir=false;
}
while(counter<=(n*n) && !changeDir)
{
newR =newR+temp.first;
newC=newC+temp.second;
if(valid(n,newR,newC) && !visited[newR][newC])
{
arr[newR][newC]=counter;
visited[newR][newC]=1;
counter++;
}
else
{
newR-=temp.first;
newC-=temp.second;
changeDir=true;
break;
}
}
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(arr[i][j]<10)
cout<<0;
cout<<arr[i][j]<<" ";
}
cout<<endl;
}
return 0;
}
这是输出，其中N = 5：
01 02 03 04 05
16 17 18 19 06
15 24 25 20 07
14 23 22 21 08
13 12 11 10 09
谢谢。