Solutions are many, while I prefer the most readable and clear one. The code is very easy to understand and learn.
Solution:
# T:O(m*n) S:O(1)
class Solution:
# @param matrix, a list of lists of integers
# @return a list of integers
def spiralOrder(self, matrix):
result = []
if matrix == []:
return result
left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1
while left <= right and top <= bottom:
for j in xrange(left, right + 1):
result.append(matrix[top][j])
for i in xrange(top + 1, bottom):
result.append(matrix[i][right])
for j in reversed(xrange(left, right + 1)):
if top < bottom:
result.append(matrix[bottom][j])
for i in reversed(xrange(top + 1, bottom)):
if left < right:
result.append(matrix[i][left])
left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1
return result
Run Time: 44 ms
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