How to find the Kth bit in the Nth binary string on LeetCode 1545?

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给定两个正整数 n 和 k，二进制字符串 Sn 是按以下方式形成的：S1=0，Si=Si-1 + 1 + reverse(invert(Si-1))，其中 + 表示连接操作，reverse(x) 返回字符串 x 的反转，invert(x) 返回字符串 x 的反转后的每个字符取反。

Description

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

分析

题目的意思是：给定变换规则，求第N次变换后第K个比特位的值。题目中的n比较小，思路比较直接，直接模拟找到答案。看了一下别人的实现，思路跟我差不多。

代码

class Solution:
def solve(self,s):
t=[]
for ch in s:
if(ch=='1'):
t.append('0')
else:
t.append('1')
t.reverse()
s=s+'1'+''.join(t)
return s
def findKthBit(self, n: int, k: int) -> str:
s='0'
for i in range(n):
s=self.solve(s)
return s[k-1]

参考文献

​​[LeetCode] Simple Python Solution with clean and understandable code​​