Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99]
)
Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.
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class Solution:
def countNumbersWithUniqueDigits(self, n):
"""
:type n: int
:rtype: int
"""
used = [False for x in range(10)]
def gen(used, n, d):
if n == d:
return 1
total = 1
startIndex = 1 if d == 0 else 0
for i in range(startIndex, 10):
if not used[i]:
used[i] = True
total += gen(used, n, d + 1)
used[i] = False
return total
return gen(used, n, 0)
s = Solution()
a = s.countNumbersWithUniqueDigits(7)
print(a)
Solution
原文地址:https://www.cnblogs.com/xiejunzhao/p/8445742.html