[leetcode] 400. Nth Digit @ python

最新推荐文章于 2020-05-28 17:58:59 发布

原创最新推荐文章于 2020-05-28 17:58:59 发布 · 191 阅读

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本文介绍了一种算法，用于找到无限整数序列1,2,3,...中的第N位数字。通过分层累加的方式确定目标数字所在层级及具体位置。

原题

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.

解法

按层次累加. 设i为i位数, 那么有以下规律

i = 1: 1, 2, ...9
i = 2: 10, 11, ...99
i = 3: 100, 101, ..., 999

我们先确定n在哪一层, 然后确定n离那一层第一个数的距离, 以确定n所在的数字, 使用divmod()方法确定n离当前层的第一个数的距离和它的index. 返回数字对应的index即可.

代码

class Solution(object):
    def findNthDigit(self, n):
        """
        :type n: int
        :rtype: int
        """
        def totalDigits(i):
            # return the total digits of i-digit number
            start = 10**(i-1)
            end = 10**i-1
            return (end-start+1)*i
        
        i = 1
        total = 0
        prev_total = 0
        # find the level of the digit that n belongs
        while True:
            total += totalDigits(i)
            if n <= total:
                break
            else:
                i += 1
                prev_total = total
        # find the distance from the first number of the level
        distance = n - prev_total-1
        div, mod = divmod(distance, i)
        num = 10**(i-1) + div
        return int(str(num)[mod])