[一起來刷leetcode吧][29]--No.400 Nth digit

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這是leetcode的第400題--Nth digit

  題目

Find the n^th 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:

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.

  思路 這道題和前面的Integer to Roman 方法類似，要根據n的範圍來劃分不同的結構，這裏就是一個數字的位數，9 , 90 * 2, 900 * 3,... 找到n在的“區間”，然後即找出這個區間的某個數，這個數的某位（digit）

  

show me the code

class Solution(object):
    def findNthDigit(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n <= 9:
            return n
        begin = 0 
        num = 0
        while n >= begin:
            begin  = 9 * (10 **num) *(num   1)
            num  = 1
        begin  -=  9 * (10**(num-1))*num
        if n == begin :
            return 9
        n -= begin
        tmp  = int((n-1)/ num)   1   int("9"*(num -1))  #floor 
        n = (n-1) % num 
        return int(str(tmp)[n])