# A straightforward O(mn) python solution without the mutating the matrix, 44ms

• Ideas:

1. use two pointers for row (`i` and `iend`) and columns (`j` and `jend`).
2. use `turn` to record the number of `while` loop so that `turn%4` indicates what direction it should be in this loop.

In this solution, every element would be visited exactly once (O(mn)) and this solution does not mutate the matrix. (many shared solutions on OJ use matrix.pop(), which would mutate the matrix )

``````    def spiralOrder(self, matrix):
if not matrix:
return []
i, iend , j, jend = 0, len(matrix)-1, 0, len(matrix[0])-1
ans, turn = [], 0
while i <= iend and j <= jend:
if turn%4 == 0: # left to right
ans += matrix[i][j:jend+1]
i += 1
elif turn%4 == 1: # top to bottom
ans += [matrix[k][jend] for k in xrange(i, iend+1)]
jend -=1
elif turn%4 == 2: # right to left
ans += matrix[iend][j:jend+1][::-1]
iend -=1
elif turn%4 == 3: # bottom to top
ans += [matrix[k][j] for k in xrange(iend, i-1,-1)]
j += 1
turn += 1
return ans
``````

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