python 生成螺旋矩阵

Posted 2021-11-26 sea-stream

 

对于任意 m*n 矩阵，将 1~m*n 的数字按照螺旋规则在矩阵中排列。

如 m=3,n=3,期望结果为：

[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]

以下代码支持方阵以及非方阵。

code:

# coding=utf-8
import numpy

flag=1
pos_x=0
pos_y=0
def inc(pos_x,pos_y,row,col):
    if(-1<pos_x<row and -1<pos_y<col):
        return True
    else:
        return False
def gen(row,col):
    global flag
    global pos_x
    global pos_y
    rowbox=[]
    for i in range(col):
        rowbox.append(0)
    data=[]
    for i in range(row):
        data.append(rowbox)
    x = numpy.array(data)
    for i in range(1,row*col+1):
        while(1):
            if(flag==1):
                if(inc(pos_x,pos_y,row,col) and x[pos_x][pos_y]==0):
                    x[pos_x][pos_y]=i
                    pos_y=pos_y+1
                    break
                else:
                    pos_y=pos_y-1
                    pos_x=pos_x+1
                    flag=2
            if(flag==2):
                if(inc(pos_x,pos_y,row,col) and x[pos_x][pos_y]==0):
                    x[pos_x][pos_y]=i
                    pos_x=pos_x+1
                    break
                else:
                    pos_x=pos_x-1
                    pos_y=pos_y-1
                    flag=3
            if(flag==3):
                if(inc(pos_x,pos_y,row,col) and x[pos_x][pos_y]==0):
                    x[pos_x][pos_y]=i
                    pos_y=pos_y-1
                    break
                else:
                    pos_y=pos_y+1
                    pos_x=pos_x-1
                    flag=4
            if(flag==4):
                if(inc(pos_x,pos_y,row,col) and x[pos_x][pos_y]==0):
                    x[pos_x][pos_y]=i
                    pos_x=pos_x-1
                    break
                else:
                    pos_y=pos_y+1
                    pos_x=pos_x+1
                    flag=1
    return x


# m*n Matrix
m=3
n=6
print(gen(m,n))

输出

[[ 1  2  3  4  5  6]
 [14 15 16 17 18  7]
 [13 12 11 10  9  8]]