問題描述：
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])
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Following the hint. Let f(n) = count of number with unique digits of length n.
f(1) = 10. (0, 1, 2, 3, …., 9)
f(2) = 9 * 9. Because for each number i from 1, …, 9, we can pick j to form a 2-digit number ij and there are 9 numbers that are different from i for j to choose from.
f(3) = f(2) * 8 = 9 * 9 * 8. Because for each number with unique digits of length 2, say ij, we can pick k to form a 3 digit number ijk and there are 8 numbers that are different from i and j for k to choose from.
Similarly f(4) = f(3) * 7 = 9 * 9 * 8 * 7….
…
f(10) = 9 * 9 * 8 * 7 * 6 * … * 1
f(11) = 0 = f(12) = f(13)….
any number with length > 10 couldn’t be unique digits number.
The problem is asking for numbers from 0 to 10^n. Hence return f(1) + f(2) + .. + f(n)
As @4acreg suggests, There are only 11 different ans. You can create a lookup table for it. This problem is O(1) in essence.
public int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
int res = 10;
int uniqueDigits = 9;
int availableNumber = 9;
while (n-- > 1 && availableNumber > 0) {
uniqueDigits = uniqueDigits * availableNumber;
res += uniqueDigits;
availableNumber--;
}
return res;
}