248. Strobogrammatic Number III
题干：
A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
Write a function to count the total strobogrammatic numbers that exist in the range of low <= num <= high.
解释：
就是把字符串翻转之后和原来相同的个数
思考：
这个题还是蛮有意思的，因为对于这种字符串，其实只要知道一半就知道全部了，而且只可能由，1，6，9，0，8这些数字构成。 其实是可以构造的，按照边长来构造，1位的有0，1，8，1.5位的的中心必须是0，1，8，边界可以是0，1，6，9，8构成 根据范围选择每一位的构造可能性，然后乘起来总选择数即可。
答案：
class Solution:
def strobogrammaticInRange(self, low, high):
len1, len2 = len(low), len(high)
result = 0
if len1 == len2:
temp = self.findStrobogrammatic(len1)
else:
for i in range(len1 + 1, len2):
result += self.process1(i)
if low == "9" * len1:
temp1 = []
else:
temp1 = self.findStrobogrammatic(len1)
if high[0] == "1" and high[1:] == "0" * (len2 - 1):
temp2 = []
else:
temp2 = self.findStrobogrammatic(len2)
temp = temp1 + temp2
index = 0
while index < len(temp):
if int(low) <= temp[index] <= int(high):
result += 1
index += 1
return result
def findStrobogrammatic(self, n):
nums = ["0", "1", "6", "8", "9", "6", "8", "9", "1", "0"]
result = []
self.process(nums, "", "", n, result)
result = list(map(int, result))
result = list(map(str, result))
index = 0
while index < len(result):
if len(result[index]) < n:
result.remove(result[index])
else:
break
result = list(map(int, result))
return result
def process(self, nums, temp1, temp2, n, result):
if n == 0:
result.append(temp1 + temp2)
elif n == 1:
result.append(temp1 + "0" + temp2)
result.append(temp1 + "1" + temp2)
result.append(temp1 + "8" + temp2)
else:
size = len(nums)
for i in range(size // 2):
self.process(nums, temp1 + nums[i], nums[size - 1 - i] + temp2, n - 2, result)
def process1(self, n):
if n == 1:
return 5
result = [4]
self.process2(n - 2, result)
return result[0]
def process2(self, n, result):
if n == 0:
return
if n == 1:
result[0] *= 3
return
else:
result[0] *= 5
self.process2(n - 2, result)
