[leetcode] 89. Gray Code @ python

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原题

The gray code is a binary numeral system where two successive values differ in only one bit.

Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.

Example 1:

Input: 2
Output: [0,1,3,2]
Explanation:
00 - 0
01 - 1
11 - 3
10 - 2

For a given n, a gray code sequence may not be uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence.

00 - 0
10 - 2
11 - 3
01 - 1
Example 2:

Input: 0
Output: [0]
Explanation: We define the gray code sequence to begin with 0.
A gray code sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
Therefore, for n = 0 the gray code sequence is [0].

解法

gray code的定义是相邻两个码之间只有一个bit不同. 首先我们推算n对应的字符串, base case是n==0, 返回[‘0’].

1位格雷码有两个码字
(n+1)位格雷码中的前2n个码字等于n位格雷码的码字，按顺序书写，加前缀0
(n+1)位格雷码中的后2n个码字等于n位格雷码的码字，按逆序书写，加前缀1

因此当n > 1时, 推导函数为

f(n) = ['0'+s for s in f(n-1)] + ['1'+s for s in f(n-1)[::-1]]

最后用map函数将列表里的二进制字符串转化为整数.

代码

class Solution(object):
    def grayCode(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        return map(lambda x: int(x, 2), self.grayBit(n))
    
        
    def grayBit(self, n):
        if n == 0:
            return ['0']
        if n == 1:
            return ['0','1']
        prev = self.grayBit(n-1)
        return ['0'+s for s in prev] + ['1'+s for s in prev[::-1]]