Easy
Strobogrammatic Number — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
class Solution(object):
lookup = {'0':'0', '1':'1', '6':'9', '8':'8', '9':'6'}
# @param {string} num
# @return {boolean}
def isStrobogrammatic(self, num):
n = len(num)
for i in xrange((n+1) / 2):
if num[n-1-i] not in self.lookup or \
num[i] != self.lookup[num[n-1-i]]:
return False
return True