思路：如果有一个数字重复出现，那么一定不是Happy Number
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
num_dict = {}
while n not in num_dict:
num_dict[n] = 1
t = 0
while n > 0:
i = n % 10
t = t + i * i
n = n / 10
n = t
if n == 1:
return True
return False
另外一种空间复杂度只需要O(1)的方法，类似于判断循环链表的快慢指针的思想
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
fast = slow = n
while True:
slow = self.getNumber(slow)
fast = self.getNumber(self.getNumber(fast))
if slow == fast:
break
if slow == 1:
return True
return False
def getNumber(self, n):
t = 0
while n > 0:
i = n % 10
t = t + i * i
n = n / 10
return t