leetcode -- Gray Code -- 数学题

class Solution:
# @return a list of integers
def grayCode(self, n):
res=[]
size=1<<n
for i in range(size):
res.append((i>>1)^i)
return res

思路2

class Solution(object):

def convertBit(self, mylist):
mylist.reverse()
res = 0
for i,c in enumerate(mylist):
res += 2 ** i * c
return res

def dfs(self, start, end, last_bit, subres, res):
if start == end + 1:
res.append(self.convertBit(subres[:]))
return
else:
if last_bit == 0:
subres[start] = 0
self.dfs(start + 1, end, 0, subres, res)
subres[start] = 1
self.dfs(start + 1, end, 1, subres, res)
else:
subres[start] = 1
self.dfs(start + 1, end, 1, subres, res)
subres[start] = 0
self.dfs(start + 1, end, 0, subres, res)

def grayCode(self, n):
"""
:type n: int
:rtype: List[int]
"""
if n == 0: return [0]
subres = [0]*n
res = []
self.dfs(0, n-1, 0, subres, res)
return res