我如何以螺旋顺序打印一个5×5的二维数组?
是否有任何公式可以让我按照螺旋顺序打印任何大小的数组?
我如何以螺旋顺序打印一个5×5的二维数组?
是否有任何公式可以让我按照螺旋顺序打印任何大小的数组?
这个想法是将矩阵视为一系列层,即右上角层和左下角层。为了螺旋打印矩阵,我们可以从这些矩阵中剥离层,打印剥离的部分,并对剩余部分进行递归调用打印。当我们没有更多的层要打印时,递归终止。
输入矩阵:
1 2 3 4
5 6 7 8
9 0 1 2
3 4 5 6
7 8 9 1
在剥掉右上角的一层后: 1 2 3 4
8
5 6 7 2
9 0 1 6
3 4 5 1
7 8 9
从子矩阵中剥离左下角层后:
6 7
5 0 1
9 4 5
3
7 8 9
从子矩阵中剥离右上方的一层后:
6 7
1
0 5
4
从子矩阵中剥离左下角层之后:
0
4
递归终止。
C 函数:
// function to print the top-right peel of the matrix and
// recursively call the print bottom-left on the submatrix.
void printTopRight(int a[][COL], int x1, int y1, int x2, int y2) {
int i = 0, j = 0;
// print values in the row.
for(i = x1; i<=x2; i++) {
printf("%d ", a[y1][i]);
}
// print values in the column.
for(j = y1 + 1; j <= y2; j++) {
printf("%d ", a[j][x2]);
}
// see if more layers need to be printed.
if(x2-x1 > 0) {
// if yes recursively call the function to
// print the bottom left of the sub matrix.
printBottomLeft(a, x1, y1 + 1, x2-1, y2);
}
}
// function to print the bottom-left peel of the matrix and
// recursively call the print top-right on the submatrix.
void printBottomLeft(int a[][COL], int x1, int y1, int x2, int y2) {
int i = 0, j = 0;
// print the values in the row in reverse order.
for(i = x2; i>=x1; i--) {
printf("%d ", a[y2][i]);
}
// print the values in the col in reverse order.
for(j = y2 - 1; j >= y1; j--) {
printf("%d ", a[j][x1]);
}
// see if more layers need to be printed.
if(x2-x1 > 0) {
// if yes recursively call the function to
// print the top right of the sub matrix.
printTopRight(a, x1+1, y1, x2, y2-1);
}
}
void printSpiral(int arr[][COL]) {
printTopRight(arr,0,0,COL-1,ROW-1);
printf("\n");
}
Python 2 代码:
import itertools
arr = [[1,2,3,4],
[12,13,14,5],
[11,16,15,6],
[10,9,8,7]]
def transpose_and_yield_top(arr):
while arr:
yield arr[0]
arr = list(reversed(zip(*arr[1:])))
print list(itertools.chain(*transpose_and_yield_top(arr)))
对于 Python 3x:
import itertools
arr = [[1,2,3,4],
[12,13,14,5],
[11,16,15,6],
[10,9,8,7]]
def transpose_and_yield_top(arr):
while arr:
yield arr[0]
arr = list(reversed(list(zip(*arr[1:]))))
print(list(itertools.chain(*transpose_and_yield_top(arr))))
rotate = lambda a: list(a[0]) + rotate(zip(*a[1:])[::-1]) if a else []
- tcharcurRow
和height-1-curRow
)的最小值,然后取(curCol
和width-1-curCol
)的最小值,并比较它们是否相同。 但是,我们需要考虑左上角的情况,即当最小值为curRow
和curCol
本身时。 在这种情况下,我们相应地减少了垂直距离。#include <stdio.h>
int shouldTurn(int row, int col, int height, int width){
int same = 1;
if(row > height-1-row) row = height-1-row, same = 0; // Give precedence to top-left over bottom-left
if(col >= width-1-col) col = width-1-col, same = 0; // Give precedence to top-right over top-left
row -= same; // When the row and col doesn't change, this will reduce row by 1
if(row==col) return 1;
return 0;
}
int directions[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
void printSpiral(int arr[4][4], int height, int width){
int directionIdx=0, i=0;
int curRow=0, curCol=0;
for(i=0; i<height*width; i++){
printf("%d ",arr[curRow][curCol]);
if(shouldTurn(curRow, curCol, height, width)){
directionIdx = (directionIdx+1)%4;
}
curRow += directions[directionIdx][0];
curCol += directions[directionIdx][1];
}
printf("\n");
}
int main(){
int arr[4][4]= {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
printSpiral(arr, 4, 4);
printSpiral(arr, 3, 4);
}
输出结果为:
1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 1 2 3 4 8 12 11 10 9 5 6 7
same
不在左上角区域时,它的值已经被更新为 0
。 - justhalfdirections
数组是一个包含四个元素的数组,每个元素都指定了坐标系中的一个方向:右(0,1),下(1,0),左(0,-1),上(-1,0)。这是因为当你向右移动时,例如,你不会改变行数(一对数字中的第一个数字为0),但你会增加列数(一对数字中的第二个数字为1)。其他三对数字同理。你可以看到这些数字在包含curRow += directions[directionIdx][0]
和curCol += directions[directionIdx][1]
的行中如何使用。请注意,当你向下移动时,行号会增加。 - justhalf以下是三种有趣的方法
Reading in spiral way can be treated like a snake moving towards boundary and turning on hitting the boundary or itself (I find it elegant and most efficient being a single loop of N iterations)
ar = [
[ 0, 1, 2, 3, 4],
[15, 16, 17, 18, 5],
[14, 23, 24, 19, 6],
[13, 22, 21, 20, 7],
[12, 11, 10, 9, 8]]
def print_spiral(ar):
"""
assuming a rect array
"""
rows, cols = len(ar), len(ar[0])
r, c = 0, -1 # start here
nextturn = stepsx = cols # move so many steps
stepsy = rows-1
inc_c, inc_r = 1, 0 # at each step move this much
turns = 0 # how many times our snake had turned
for i in range(rows*cols):
c += inc_c
r += inc_r
print ar[r][c],
if i == nextturn-1:
turns += 1
# at each turn reduce how many steps we go next
if turns%2==0:
nextturn += stepsx
stepsy -= 1
else:
nextturn += stepsy
stepsx -= 1
# change directions
inc_c, inc_r = -inc_r, inc_c
print_spiral(ar)
输出:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
A recursive approach would be to print outer layer and call same function for inner rectangle e.g.
def print_spiral(ar, sr=0, sc=0, er=None, ec=None):
er = er or len(ar)-1
ec = ec or len(ar[0])-1
if sr > er or sc > ec:
print
return
# print the outer layer
top, bottom, left, right = [], [], [], []
for c in range(sc,ec+1):
top.append(ar[sr][c])
if sr != er:
bottom.append(ar[er][ec-(c-sc)])
for r in range(sr+1,er):
right.append(ar[r][ec])
if ec != sc:
left.append(ar[er-(r-sr)][sc])
print " ".join([str(a) for a in top + right + bottom + left]),
# peel next layer of onion
print_spiral(ar, sr+1, sc+1, er-1, ec-1)
最后,这里有一个小片段可以实现它,虽然不是很高效但很有趣 :),基本上它会打印顶行,并逆时针旋转整个矩形并重复。
def print_spiral(ar):
if not ar: return
print " ".join(str(a) for a in ar[0]),
ar = zip(*[ reversed(row) for row in ar[1:]])
print_spiral(ar)
public class circ {
public void get_circ_arr (int n,int [][] a)
{
int z=n;
{
for (int i=0;i<n;i++)
{
for (int l=z-1-i;l>=i;l--)
{
int k=i;
System.out.printf("%d",a[k][l]);
}
for (int j=i+1;j<=z-1-i;j++)
{
int k=i;
{
System.out.printf("%d",a[j][k]);
}
}
for (int j=i+1;j<=z-i-1;j++)
{
int k=z-1-i;
{
System.out.printf("%d",a[k][j]);
}
}
for (int j=z-2-i;j>=i+1;j--)
{
int k=z-i-1;
{
System.out.printf("%d",a[j][k]);
}
}
}
}
}
}
当我学习 Ruby 时,我非常困扰于这个问题。以下是我所能做到的最好的办法:
def spiral(matrix)
matrix.empty? ? [] : matrix.shift + spiral(matrix.transpose.reverse)
end
JavaScript解决方案:
var printSpiral = function (matrix) {
var i;
var top = 0;
var left = 0;
var bottom = matrix.length;
var right = matrix[0].length;
while (top < bottom && left < right) {
//print top
for (i = left; i < right; i += 1) {
console.log(matrix[top][i]);
}
top++;
//print right column
for (i = top; i < bottom; i += 1) {
console.log(matrix[i][right - 1]);
}
right--;
if (top < bottom) {
//print bottom
for (i = right - 1; i >= left; i -= 1) {
console.log(matrix[bottom - 1][i]);
}
bottom--;
}
if (left < right) {
//print left column
for (i = bottom - 1; i >= top; i -= 1) {
console.log(matrix[i][left]);
}
left++;
}
}
};
保持简单 -->
public class spiralMatrix {
public static void printMatrix(int[][] matrix, int rows, int col)
{
int rowStart=0;
int rowEnd=rows-1;
int colStart=0;
int colEnd=col-1;
while(colStart<=colEnd && rowStart<=rowEnd)
{
for(int i=colStart;i<colEnd;i++)
System.out.println(matrix[rowStart][i]);
for(int i=rowStart;i<rowEnd;i++)
System.out.println(matrix[i][colEnd]);
for(int i=colEnd;i>colStart;i--)
System.out.println(matrix[rowEnd][i]);
for(int i=rowEnd;i>rowStart;i--)
System.out.println(matrix[i][colStart]);
rowStart++;
colEnd--;
rowEnd--;
colStart++;
}
}
public static void main(String[] args){
int[][] array={{1,2,3,4},{5,6,7,8}};
printMatrix(array,2,4);
}
}
challenge = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
def spiral_elements(array):
while array:
first_row, *rest = array
yield from first
array = tuple(reversed(tuple(zip(*rest))))
result = list(spiral_elements(challenge))
print(result)
# [1, 2, 3, 6, 9, 8, 7, 4, 5]
#include <stdio.h>
#define N_ROWS 5
#define N_COLS 3
void print_spiral(int a[N_ROWS][N_COLS])
{
enum {up, down, left, right} direction = right;
int up_limit = 0,
down_limit = N_ROWS - 1,
left_limit = 0,
right_limit = N_COLS - 1,
downcount = N_ROWS * N_COLS,
row = 0,
col = 0;
while(printf("%d ", a[row][col]) && --downcount)
if(direction == right)
{
if(++col > right_limit)
{
--col;
direction = down;
++up_limit;
++row;
}
}
else if(direction == down)
{
if(++row > down_limit)
{
--row;
direction = left;
--right_limit;
--col;
}
}
else if(direction == left)
{
if(--col < left_limit)
{
++col;
direction = up;
--down_limit;
--row;
}
}
else /* direction == up */
if(--row < up_limit)
{
++row;
direction = right;
++left_limit;
++col;
}
}
void main()
{
int a[N_ROWS][N_COLS] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
print_spiral(a);
}