这个题目是没有官方Sol的。用循环是不可以解出的(Memory or Time exceeded)，罕见的必须要用递归的题目，而且递归的写法也很优雅

We build a table of n rows (1-indexed). We start by writing 0 in the 1st row. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.

• For example, for n = 3, the 1st row is 0, the 2nd row is 01, and the 3rd row is 0110.

Given two integer n and k, return the kth (1-indexed) symbol in the nth row of a table of n rows.

Example 1:

Input: n = 1, k = 1
Output: 0
Explanation: row 1: 0

Example 2:

Input: n = 2, k = 1
Output: 0
Explanation: 
row 1: 0
row 2: 01

Example 3:

Input: n = 2, k = 2
Output: 1
Explanation: 
row 1: 0
row 2: 01

Constraints:

• 1 <= n <= 30
• 1 <= k <= 2n - 1

  class Solution:
      def kthGrammar(self, n, k):
          if n==1:
              return 0
          else:
              if k<=(1<<(n-1))/2:
                  return self.kthGrammar(n-1,k)
              else:
                  return 1-self.kthGrammar(n-1,k-(1<<(n-1))/2)