Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] Output: [1,2,3,4,8,12,11,10,9,5,6,7]
My code:
class Solution:
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if len(matrix)==0:
return []
direction=0
self.record=[]
tmp_record=matrix[0]
if len(tmp_record)==0:
return []
self.check(matrix,0,len(matrix),0,len(tmp_record),direction)
return self.record
def check(self,matrix,i,j,m,n,direction):
#print('i %d, j %d'%(i,j))
if i>=j or m>=n:
return
if j-i==1:
for k in range(m,n):
self.record.append(matrix[i][k])
return
if n-m==1:
for k in range(i,j):
self.record.append(matrix[k][m])
return
if direction==0:
for k in range(m,n):
self.record.append(matrix[i][k])
self.check(matrix,i,j,m,n,direction+1)
elif direction==1:
for k in range(i+1,j):
self.record.append(matrix[k][n-1])
self.check(matrix,i,j,m,n,direction+1)
elif direction==2:
#print('m %d n %d'%(m,n))
for k in range(n-2,m-1,-1):
self.record.append(matrix[j-1][k])
self.check(matrix,i,j,m,n,direction+1)
elif direction==3:
for k in range(j-2,i,-1):
self.record.append(matrix[k][m])
direction=0
self.check(matrix,i+1,j-1,m+1,n-1,direction)