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题目：

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:3Output:3

Example 2:

Input:11Output:0Explanation: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

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   1-92   10-993   100-9994   1000-99995   10000-999996   100000-9999997   1000000-99999998   10000000-999999999   100000000-99999999

然后分区间求值即可。

方法一：性能56ms

class Solution(object):    def findNthDigit(self, n):        """        :type n: int        :rtype: int        """        for i in range(9):            d = 9 * pow(10, i)            if n <= d * (i + 1): break            n -= d * (i+1)        n -= 1        return int(str(pow(10, i) + n/(i+1))[n % (i+1)])

方法二：性能38ms

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

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