给定一个包含 m x n 个元素的矩阵(m 行, n 列),请按照顺时针螺旋顺序,返回矩阵中的所有元素。
示例1:
输入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
示例2:
输入:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]
1 思路:
- 从[i,i]坐标开始按照顺时针输出每一层的数字
- 一共循环的次数$ n = \left \lceil min(row,col)/2 \right \rceil $
对于一维矩阵直接输出,
思路比较简单就是索引比较麻烦。运行时间比较长。时间复杂度O(n)。n是col*row.
import numpy as np
class Solution:
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
def toouter(matrix,indexs):
n,m = indexs
outer = []
p,q = len(matrix),len(matrix[0])
for i in range(m,q-m):
outer.append(matrix[n][i])
for i in range(n+1,p - n):
outer.append(matrix[i][q-m-1])
if p-n-1 > n:
for i in range(q-m-2,m-1,-1):
outer.append(matrix[p-n-1][i])
if q-m-1 > m:
for i in range(p - n -2,n,-1):
outer.append(matrix[i][m])
return outer
if len(np.array(matrix).shape) == 2:
size = len(matrix),len(matrix[0])
minnum = size[0] if size[0] < size[1] else size[1]
indexnum = int(np.ceil(minnum / 2))
outarray = []
for i in range(indexnum):
outarray.extend(toouter(matrix,(i,i)))
return outarray
else:
return matrix