Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
解法1:
class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
# 1-9=1-9
# 10=1
# 11=2
# 12=3 ...
# 18=9
# 19=>1
# 20=>2
# 21=>3
# 99=>9
# 100=>1
# 101=>2
# 999=>9
def sum_digits(n):
ans = 0
while n:
ans += n%10
n /= 10
return ans ans = num
while ans > 9:
ans = sum_digits(ans)
return ans
观察发现是一个循环数组：
class Solution(object):
def addDigits(self, num):
"""
:type num: int
:rtype: int
"""
# 1-9=1-9
# 10=1
# 11=2
# 12=3 ...
# 18=9
# 19=>1
# 20=>2
# 21=>3
# 99=>9
# 100=>1
# 101=>2
# 999=>9
if num == 0: return 0
return 9 if num % 9 == 0 else num % 9