题目描述:
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.
题目大意:
给定一个无穷整数序列1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... 求序列的第n位数字。
注意:
n是正数并且范围在32位带符号整数之内(n < 2^31)
解题思路:
将整数序列划分为下列区间:
1 1-9 2 10-99 3 100-999 4 1000-9999 5 10000-99999 6 100000-999999 7 1000000-9999999 8 10000000-99999999 9 100000000-99999999
然后分区间求值即可。
Python代码:
class Solution(object):
def findNthDigit(self, n):
"""
:type n: int
:rtype: int
"""
for i in range(9):
d = 9 * 10 ** i
if n <= d * (i + 1): break
n -= d * (i + 1)
n -= 1
return int(str(10**i + n / (i + 1))[n % (i + 1)])
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