[Leetcode][Python]54: Spiral Matrix
# -*- coding: utf8 -*-
‘‘‘
__author__ = ‘[email protected]‘
54: Spiral Matrix
https://leetcode.com/problems/spiral-matrix/
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].
=== Comments by Dabay===
一圈一圈的处理,一共处理n/2圈。
注意一些细节,比如圈数,还有同一圈不要重复。
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class Solution:
# @param matrix, a list of lists of integers
# @return a list of integers
def spiralOrder(self, matrix):
if len(matrix) == 0:
return []
if len(matrix) == 1:
return matrix[0]
if len(matrix[0]) == 1:
return [i[0] for i in matrix]
height = len(matrix)
width = len(matrix[0])
res = []
n = 0
while n < (min(height, width)+1)/2:
for i in xrange(n, width-n):
res.append(matrix[n][i])
for i in xrange(n+1, height-n):
res.append(matrix[i][width-1-n])
if n < height-n-1:
for i in reversed(xrange(n, width-n-1)):
res.append(matrix[height-n-1][i])
for i in reversed(xrange(n+1, height-n-1)):
res.append(matrix[i][n])
n += 1
return res
def main():
sol = Solution()
matrix = [
[2, 3]
]
print sol.spiralOrder(matrix)
if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)
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