题目
即判断一个数是不是反过来之后还和以前是一样的
解法如下,提供了两种解法,一种是直接判断,一种是先建立一个字典
def isStrobogrammatic(num):
num = str(num)
left, right = 0, len(num)-1
while (left<=right):
if num[left]==num[right]:
if num[left] != "1" and num[left] !="0" and num[left] != "8":
return False
else:
if (num[left] !="6" or num[right] != "9") and (num[left] != "9" or num[right] != "6"):
return False
left += 1
right -= 1
return True
import sys
ss = int(sys.argv[1])
print(ss, isStrobogrammatic(ss))
## 法二,建立一个字典
dic = {'0':'0', '1':'1', '8':'8', '6':'9', '9':'6'}
def isStrobogrammatic2(num):
num = str(num)
for i in range(len(num)//2+1):
if num[i] in dic.keys():
if dic[num[i]] != num[len(num)-1-i]:
return False
else:
return False
return True
print(ss, isStrobogrammatic2(ss))





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