寫在最前面：

為什麼我覺得這道簡單的題好難啊...感覺自己是個數學渣渣

純原創，不會有人和我一樣奇葩的思路的...

leetcode【400】Nth Digit

Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 231).

Example 1:

Input:
3

Output:
3
 

Example 2:

Input:
11

Output:
0

Explanation:
The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.

求出幾位數

在這個幾位數的第幾個

屬於哪個數字

屬於哪一個數字的第幾個字元

自己領悟吧~

class Solution:
    def findNthDigit(self, n):
        """
        :type n: int
        :rtype: int
        """
        digit = 1
        while n > digit*9*10**(digit-1):
            n -= digit*9*10**(digit-1)
            digit += 1
        a = n // digit
        b = n % digit
        if a == 0:
            c = 10 ** (digit-1) + a
        else:
            c = 10 ** (digit-1) + a -1
        if b == 0:
            return int(str(c)[digit-1])
        else:
            return int((str(c+1)[b-1]))