难度: Hard
原题连接
内容描述
Start from integer 1, remove any integer that contains 9 such as 9, 19, 29...
So now, you will have a new integer sequence: 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, ...
Given a positive integer n, you need to return the n-th integer after removing. Note that 1 will be the first integer.
Example 1:
Input: 9
Output: 10
Hint: n will not exceed 9 x 10^8
思路 1 - 时间复杂度: O(1)- 空间复杂度: O(1)******
Let's write the first numbers and try to notice a pattern. Those numbers are:
1, 2, 3, 4, 5, 6, 7, 8,
10, 11, 12, 13, 14, 15, 16, 17, 18,
20, 21, 22, 23, 24, 25, 26, 27, 28,
...
80, 81, 82, 83, 84, 85, 86, 87, 88,
100, 101, 102, ...
These numbers look exactly like all base-9 numbers!
Indeed, every base-9 number is a number in this sequence,
and every number in this sequence is a base-9 number.
Both this sequence and the sequence of all base-9 numbers are in increasing order.
The answer is therefore just the n-th base-9 number.
递归
class Solution:
def newInteger(self, n):
"""
:type n: int
:rtype: int
"""
return n if n < 9 else 10 * self.newInteger(n // 9) + n % 9思路 2 - 时间复杂度: O(1)- 空间复杂度: O(1)******
class Solution:
def newInteger(self, n):
"""
:type n: int
:rtype: int
"""
res = ''
while n:
res = str(n % 9) + res
n //= 9
return int(res)what if we are not removing 9 but removing 1, 2, 3, ... , 8?
思路 1 - 时间复杂度: O(1)- 空间复杂度: O(1)******
参考Alternative solution applicable to the general case
class Solution:
def newInteger(self, n):
"""
:type n: int
:rtype: int
"""
def count(m, d): # count how many number has 'd' in it in range [1, m]
res, p, q = 0, 1, 1
n = m
while n >= 10:
n //= 10
p *= 10
q *= 9
n = m
while n >= d:
a = n // p
res += a * (p - q)
if a == d:
res += n % p + 1
break
elif a > d:
res += q
n %= p
p //= 10
q //= 9
return res
def helper(n, remove):
pre = cur = 0
cur = count(n + cur, remove)
while pre != cur:
pre = cur
cur = count(n + cur, remove)
return int(min(n + cur, float('inf')))
return helper(n, 9)测试了一下remove 7 的话，第70个 number 是88，测试通过！！