Python使用迭代器打印螺旋矩阵的思路及代码示例

import itertools 
def spiral(n,m): 
  _status = itertools.cycle(['right','down','left','up'])#用于状态周期性的切换 
  _movemap = { 
    'right':(1,0), 
    'down':(0,1), 
    'left':(-1,0), 
    'up':(0,-1), 
  } 
  pos2no = dict.fromkeys([(x,y) for x in range(n) for y in range(m)]) 
  _pos = (0,0) 
  _st = next(_status) 
  for i in range(1,n*m+1): 
    _oldpos = _pos 
    _pos = tuple(map(sum,zip(_pos,_movemap[_st])))#根据状态进行移动 
    if (_pos not in pos2no) or (pos2no[_pos]):#当超出范围或遇到障碍时切换方向 
      _st = next(_status) 
      _pos = tuple(map(sum,zip(_oldpos,_movemap[_st]))) 
    pos2no[_oldpos] = i 
  return pos2no 
 
def display_spiral(n,m): 
  pos2no = spiral(n,m) 
  for i in range(m): 
    for j in range(n): 
      print pos2no[(j,i)],'\t', 
    print '\n' 
  print '-'*30 
 
display_spiral(4,4) 
display_spiral(5,4)