【leetcode】1545. Find Kth Bit in Nth Binary String

题目如下：

Given two positive integers n and k, the binary string  Sn is formed as follows:

• S1 = "0"
• Si = Si-1 + "1" + reverse(invert(Si-1)) for i > 1

Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).

For example, the first 4 strings in the above sequence are:

• S1 = "0"
• S2 = "011"
• S3 = "0111001"
• S4 = "011100110110001"

Return the kth bit in Sn. It is guaranteed that k is valid for the given n.

Example 1:

Input: n = 3, k = 1
Output: "0"
Explanation: S3 is "0111001". The first bit is "0".

Example 2:

Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001". The 11th bit is "1".

Example 3:

Input: n = 1, k = 1
Output: "0"

Example 4:

Input: n = 2, k = 3
Output: "1"

Constraints:

• 1 <= n <= 20
• 1 <= k <= 2n - 1

解题思路：n最大才20，先把Sn算出来即可。

代码如下：

class Solution(object):
    def findKthBit(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: str
        """
        bs = '0'

        def reverse_invert(input):
            output = ''
            for i in input:
                output += '1' if i == '0' else '0'
            return output[::-1]

        for i in range(1,n+1):
            bs = bs + '1' + reverse_invert(bs)
        return bs[k-1]